Giải hpta)$\left\{{}\begin{matrix}\dfrac{2x}{x 1} \dfrac{y}{y 1}=2\\\dfrac{x}{x 1} \dfrac{3}{y 1}=-1\end{matrix}\right.$b)\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y 12}=1\\\dfrac{x}{y 12}-\df...

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 11 2023

a: ĐKXĐ: x<>-1 và y<>-1

$\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{2x+2-2}{x+1}+\dfrac{y+1-1}{y+1}=2\\\dfrac{x+1-1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}2-\dfrac{2}{x+1}+1-\dfrac{1}{y+1}=2\\1-\dfrac{1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=2-3=-1\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=1+1=2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=-1\\\dfrac{2}{x+1}-\dfrac{6}{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y-1}=3\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}y-1=-\dfrac{7}{3}\\\dfrac{1}{x+1}-3:\dfrac{-7}{3}=2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}+3\cdot\dfrac{3}{7}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}=2-\dfrac{9}{7}=\dfrac{5}{7}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x+1=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\left(nhận\right)$

b: ĐKXĐ: y<>0 và y<>-12

$\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y}-\dfrac{x}{y+12}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\cdot\dfrac{x}{y+12}=3\left(vôlý\right)\\\dfrac{x}{y}-\dfrac{x}{y+12}=1\end{matrix}\right.$

Vậy: $\left(x,y\right)\in\varnothing$

d: ĐKXĐ: $\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.$

$\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.$

=>$\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)$

Đúng(2)

D

DUTREND123456789

26 tháng 11 2023

Giải...

Giải hệ phương trình : a) $\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.$ b) $\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.$ c) $\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.$ d) $\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$ e)...

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 11 2023

a: ĐKXĐ: $\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.$

$\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.$

=>$\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.$

=>$\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.$(nhận)

b: ĐKXĐ: $\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.$

$\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x-y+2=1\\3+\dfrac{2}{x+y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x+y-1}=1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x-y=-1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.$

=>$\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.$(nhận)

c:

ĐKXĐ: $\left\{{}\begin{matrix}y< >2x\\y< >-x\end{matrix}\right.$

$\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x+y=3\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=6\\2x-y=3\end{matrix}\right.$

=>x=2 và y=2x-3=4-3=1(nhận)

d:ĐKXĐ: $\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.$

$\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.$

=>$\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)$

e:

ĐKXĐ: $x\ne\pm2y$

$\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.$

=>$\left\{{}\begin{matrix}-\dfrac{6}{x+2y}=5\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.$

=>$\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)$

Đúng(2)

H

Học24

8 tháng 1 2018

1

T

thuongnguyen

8 tháng 1 2018

$a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.$ (1)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = $\dfrac{1}{x}$ ; b = $\dfrac{1}{y}$

Thay vào (1) ta được :

$\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.$

Vậy S = {($\dfrac{1}{5};-\dfrac{1}{8}$)}

$b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.$ (2)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = 1/x ; b = 1/y

Thay vào (2) ta được : $\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.$

Vậy S = {($\dfrac{3}{4};-3$ )}

c) $\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.$

ĐK xác định : x≠0 ; y ≠0

Áp dụng quy tác cộng đại số ta có :

$\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.$

Vậy S = {($\dfrac{3}{28};\dfrac{3}{13}$)}

d) $\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$

ĐK xác định : x≠0 ; y≠0

áp dụng quy tắc cộng đại số ta có :

$\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.$

Vậy S = {($-\dfrac{5}{11};-\dfrac{5}{9}$)}

e) ĐK xác định x≠0 ; y≠0

$\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.$ <=>$\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.$

Vậy S={($\dfrac{17}{2};-\dfrac{17}{22}$)}

TT

Trần Thị Mỹ Trinh

30 tháng 11 2021

giải các hệ phương trình

a)$\left\{{}\begin{matrix}x^2+y^2=1\\x^3+y^3=1\end{matrix}\right.$ b) $\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{5}{12}\\x^2+y^2=1\end{matrix}\right.$

c)$\left\{{}\begin{matrix}x^2-xy+y^2=3\\2x^2-xy+3y^2=12\end{matrix}\right.$

0

NN

nguyen ngoc son

27 tháng 12 2021

giai hpt

a.$\left\{{}\begin{matrix}x=y+4\\2x+3=0\end{matrix}\right.$

b.$\left\{{}\begin{matrix}2x+y=7\\3y-x=7\end{matrix}\right.$

c.$\left\{{}\begin{matrix}5x+y=3\\-x-\dfrac{1}{5}y=\dfrac{-3}{5}\end{matrix}\right.$

d.$\left\{{}\begin{matrix}3x-5y=-18\\x-5=2y\end{matrix}\right.$

1

RH

Rin Huỳnh

27 tháng 12 2021

$a) \begin{cases}x=y+4\\2x+3=0\end{cases}\Leftrightarrow\begin{cases}x = y + 4\\2x = -3\end{cases}\Leftrightarrow\begin{cases}\dfrac{-3}{2} = y + 4\\x = \dfrac{-3}{2}\end{cases}\Leftrightarrow\begin{cases}y = \dfrac{-11}{2}\\x = \dfrac{-3}{2}\end{cases}\\b) \begin{cases}2x + y = 7\\3y - x = 7\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\6y - 2x = 14\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\7y = 21\end{cases}\Leftrightarrow\begin{cases}2x + 3 = 7\\y = 3\end{cases}\Leftrightarrow\begin{cases}x=2\\y=3\end{cases}\\ c) \begin{cases} 5x + y = 3 \\ -x - \dfrac{1}{5}y=\dfrac{-3}{5} \end{cases} \Leftrightarrow \begin{cases} 5x + y = 3 \\ 5x + y = 3 \end{cases} (luôn\ đúng) \Leftrightarrow Phương\ trình\ vô\ số\ nghiệm \\d) \begin{cases} 3x - 5y = -18 \\ x - 5 = 2y \end{cases} \Leftrightarrow \begin{cases} 3x - 5y = -18 \\ 3x - 6y = 15 \end{cases} \Leftrightarrow \begin{cases} x - 5 = 2.(-33)\\ y = -13 \end{cases} \Leftrightarrow \begin{cases}x = -61\\y=-33 \end{cases} $

Đúng(1)

TT

Thầy Tùng Dương

VIP

28 tháng 1 2021

Câu hỏi hay

Giải các hệ phương...

31

NT

Nguyễn Thị Kim Thoa 1977, GV Địa–Ho...

3 tháng 2 2021

Phương trình đâu bạn ?

VD

Vũ Diệu Linh

4 tháng 2 2021

$x = 144,$ $y = 36$ .

NN

nguyễn ngọc trang

25 tháng 7 2018

a)$\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.$

b)$\left\{{}\begin{matrix}\dfrac{8}{x}-\dfrac{1}{y+12}=1\\\dfrac{1}{x}+\dfrac{5}{y+12}=36\end{matrix}\right.$

c)$\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=1\end{matrix}\right.$

1

MP

Mysterious Person

17 tháng 8 2018

mk lm 1 bài còn lại bn lm tương tự nha :

a) điều kiện xác định : $x\ge0;y\ge1$

đặc $a=\sqrt{x};b=\sqrt{y-1}$

$\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}a+2b=5\\4a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.$

ta có : $a=1\Rightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tmđk\right)$ ; $b=2\Rightarrow\sqrt{y-1}=2\Leftrightarrow y=5\left(tmđk\right)$

vậy phương trình có nghiệm duy nhất $\left(x;y\right)=\left(1;5\right)$

b) bn đặc : $a=\dfrac{1}{x};b=\dfrac{1}{y+12}$

c) bn đặc : $a=\dfrac{x}{x+1};b=\dfrac{y}{y+1}$

nhớ điều kiện nha

Giải hệ phương trình: a) $\left\{{}\begin{matrix}2x-y=3\\x+2y=-1\end{matrix}\right.$ b) $\left\{{}\begin{matrix}\dfrac{3}{2}x-y=\dfrac{1}{2}\\3x-2y=1\end{matrix}\right.$ c) $\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.$ d) $\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.$ e)...

QC

Qúy Công Tử

31 tháng 1 2019

4

NT

Nguyễn Thành Trương

1 tháng 2 2019

$a)\left\{{}\begin{matrix}2x-y=3\\x+2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=3\\2x+4y=-2\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\2x+4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.$

Vậy nghiệm hệ phương trình là (1; -1)

$b)\left\{{}\begin{matrix}\dfrac{3}{2}x-y=\dfrac{1}{2}\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=1\\3x-2y=1\end{matrix}\right.\Leftrightarrow0x-0y=0\left(VSN\right)$

Vậy hệ phương trình vô số nghiệm

NT

Nguyễn Thành Trương

1 tháng 2 2019

$c)\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-15y-12\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}5x-3x+10y=-1\\2x-3x+15y=-12-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+15y=-16\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-2x+30y=-32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40y=-33\\-2x+30y=-32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=\dfrac{29}{8}\end{matrix}\right.$

Vậy nghiệm hệ phương trình là $\left(\dfrac{29}{8};-\dfrac{33}{40}\right)$

Giải các hệ phương trình sau bằng cách đặt ẩn số phụ: 1) $\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.$ 2) $\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.$ 3) $\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.$ 4)...

NT

nguyen thi khanh nguyen

17 tháng 1 2018