K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

AH
Akai Haruma
Giáo viên
14 tháng 8 2021

Lời giải:
PT \(\Leftrightarrow (\frac{x+1}{11}-1)-(\frac{2x-5}{15}-1)=(\frac{3x-47}{17}+1)-(\frac{4x-59}{19}+1)\)

\(\Leftrightarrow \frac{x-10}{11}-\frac{2(x-10)}{15}=\frac{3(x-10)}{17}-\frac{4(x-10)}{19}\)

\(\Leftrightarrow (x-10)(\frac{1}{11}+\frac{4}{19}-\frac{2}{15}-\frac{3}{17})=0\)

\(\Leftrightarrow x-10=0\Leftrightarrow x=10\)

 

20 tháng 4 2021

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

 

28 tháng 9 2021

h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)

Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)

\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)

Thay a,b:

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)

9 tháng 2 2023

ĐKXĐ : \(x\inℝ\)

Ta có : \(\dfrac{x^2+4x+5}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)

\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)

\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)

Phương trình (1) <=> 5(x2 - 3x + 5) = 3(x2 - x + 5)

<=> 2x2 - 12x + 10 = 0

<=> x2 - 6x + 5 = 0

<=> (x - 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Tập nghiệm \(S=\left\{0;1;5\right\}\)

 

a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)

31 tháng 5 2021

\(\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{4}{11}\\\dfrac{2x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)(x,y\(\ne0\))<=>\(\left\{{}\begin{matrix}\dfrac{4}{y}+\dfrac{3}{x}=\dfrac{4}{11}\\\dfrac{2}{y}+\dfrac{1}{x}=\dfrac{4}{5}\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a\)

\(\dfrac{1}{y}=b\)

=>\(\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\3a+6b=\dfrac{12}{5}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}-2b=-\dfrac{112}{55}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}b=\dfrac{56}{55}\\a=\dfrac{-68}{55}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=a=-\dfrac{68}{55}\\\dfrac{1}{y}=b=\dfrac{56}{55}\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{-55}{68}\left(TM\right)\\y=\dfrac{55}{56}\left(TM\right)\end{matrix}\right.\)

vậy...

Bạn xem hình tham khảo nhé

undefinedundefined

16 tháng 2 2022

Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)

\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)

Lấy (2)-(1) ta đc:

\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(x-1\right)\\1=y+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

16 tháng 2 2022

Với \(x\ne1;y\ne-2\)

hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}2x-2=x\\\dfrac{1}{y+2}=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y+2=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)

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

2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)

\(=3\sqrt{x}\)