A = \(\dfrac{5xy^2-3z}{3xy} \dfrac{4x^2y 3z}{3xy}\)B = \(\dfrac{3y 5}{y-1} \dfrac{-y^2-4y}{1-y} \dfrac{y^2 y 7}{y-1}\)C = \(\dfrac{6x}{x^2-9} \dfrac{5x}{x-3} \dfrac{x}{x 3}\)D = \(\dfrac{1-3x}{2x} \df...

1

NL

Nguyễn Lê Phước Thịnh

18 tháng 10 2021

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

KS

Kudo Shinichi

18 tháng 10 2017

a,\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{+6x}\)

b, \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)

c,\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\left(x,y,zkhac0\right)\)

d, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)

4

NQ

Nguyễn Quang Ngọc Trác

22 tháng 10 2017

a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x

Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y

= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y

Ta lại có : 1 + 4y/24 = 1+4y / 9+3y

=> 24=9+3y => 15=3y => y=5

Vậy y=5

Nhớ like

NQ

Nguyễn Quang Ngọc Trác

22 tháng 10 2017

b, 1+3y/12 = 1+5y/5x = 1+7y/4x

Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x

= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x

Ta lại có: 1+5y / 5x = 1+5y / 6+2x

=> 5x = 6+2x => 3x = 6 => x=2

Vậy x =2

DA

Đức Anh Ramsay

17 tháng 2 2021

a) \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

b)\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

2

NP

Nguyễn Phương Linh

17 tháng 2 2021

a) ĐKXĐ: \(x\ne0\)

\(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{2\left(4x+1\right)+2x-3}{6x}\)

\(=\dfrac{10x-1}{6x}\)

b) ĐKXĐ: \(x,y\ne0\)

\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

\(=\dfrac{\left(x-y\right).\left(x+y\right)}{6x^2y^2}.\dfrac{3xy}{x+y}\)

\(=\dfrac{x-y}{2xy}\)

Đúng(3)

NL

Nguyễn Lê Phước Thịnh

17 tháng 2 2021

a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{2\left(4x+1\right)}{6x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{8x+2+2x-3}{6x}\)

\(=\dfrac{10x-1}{6x}\)

b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\cdot\dfrac{3xy}{x+y}\)

\(=\dfrac{x-y}{2xy}\)

tìm x;y;z a) \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\) b) \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\) c) \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\) và \(2x^2+2y^2-z^2\) d) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\) ...

GN

Giấu- Ñỗißuồn- VàoMưą-

27 tháng 11 2017

0

D

dream

17 tháng 9 2021

1, x : y : z = 2 : 3 : 4 và x + y + z = 18

2, \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}\) và 4x - 3y - 2z = 81

3, \(\dfrac{x}{3}=\dfrac{y}{2};\) 4y = 3z và x + y +z = 46

4, 5x = 3y; \(\dfrac{y}{z}=\dfrac{3}{2}\) và 2x + 3y -4z =34

2

MH

Minh Hiếu

17 tháng 9 2021

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

Đúng(2)

NH

Nguyễn Hoàng Minh

17 tháng 9 2021

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

Đúng(4)

ND

nguyễn đăng

19 tháng 12 2021

quy đồng các phân thức sau

a,\(\dfrac{x+1}{x-1};\dfrac{x-1}{x+1};\dfrac{4}{1-x^2}\)

b,\(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3};\dfrac{x}{y^2xy}\)

c,\(\dfrac{4x}{x-2};\dfrac{3x}{x-2};\dfrac{12x}{x^2-4}\)

d,\(\dfrac{7}{x};\dfrac{x}{x+6};\dfrac{36}{x^2+6x}\)

1

NH

Nguyễn Hoàng Minh

19 tháng 12 2021

\(a,\left(1\right)=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(2\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(3\right)=\dfrac{-4}{\left(x-1\right)\left(x+1\right)}\\ b,\left(1\right)=\dfrac{x^4y^3}{xy^3\left(x-y\right)^3};\left(2\right)=\dfrac{x\left(x-y\right)^3}{xy^3\left(x-y\right)^3}\\ c,\left(1\right)=\dfrac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(2\right)=\dfrac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(3\right)=\dfrac{12x}{\left(x-2\right)\left(x+2\right)}\\ d,\left(1\right)=\dfrac{7\left(x+6\right)}{x\left(x+6\right)};\left(2\right)=\dfrac{x^2}{x\left(x+6\right)};\left(3\right)=\dfrac{36}{x\left(x+6\right)}\)

Đúng(1)

D

dream

15 tháng 9 2021

1, \(\dfrac{x}{3}=\dfrac{y}{2}\); \(\dfrac{x}{5}=\dfrac{x}{7}\) và x +2y = -112

2, 2x = 3y; 5y = 7z và 3x - 7y + 5z = 30

3, \(\dfrac{x}{y}=\dfrac{10}{9};\) \(\dfrac{y}{z}=\dfrac{3}{4}\) và x +2y - 3z = -48

Làm giúp, cần gấp

4

MH

Minh Hiếu

15 tháng 9 2021

\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)

\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)

⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)

⇒x=42,y=28,z=20

Đúng(1)

MH

Minh Hiếu

15 tháng 9 2021

\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)

\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)

⇒x=48,y=32,z=336/5

1, Thực hiện phép tính: a, \(\dfrac{1-3x}{2}+\dfrac{x+3}{2}\) b, \(\dfrac{2\left(x+y\right)\left(x-y\right)}{x}-\dfrac{-2y^2}{x}\) c, \(\dfrac{3x+1}{x+y}-\dfrac{2x-3}{x+y}\) d, \(\dfrac{xy}{2x-y}-\dfrac{x^2-1}{y-2x}\) e, \(\dfrac{4x-1}{3x^2y}-\dfrac{7x-1}{3x^2y}\) 2, Thực hiện phép tính: a, \(\dfrac{1}{x}.\dfrac{6x}{y}\) b, \(\dfrac{2x^2}{y}.3xy^2\) c, \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\) d, \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\) e,...

AV

Ánh Vũ Ngọc

15 tháng 11 2017

1

NN

Nguyễn Nam

15 tháng 11 2017

2)

a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)

\(=\dfrac{6x}{xy}\)

\(=\dfrac{6}{y}\)

b) \(\dfrac{2x^2}{y}.3xy^2\)

\(=\dfrac{2x^2.3xy^2}{y}\)

\(=\dfrac{6x^3y^2}{y}\)

\(=6x^3y\)

c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)

\(=\dfrac{15x.2y^2}{7y^3.x^2}\)

\(=\dfrac{30xy^2}{7x^2y^3}\)

\(=\dfrac{30}{7xy}\)

d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)

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

\(=\dfrac{2y}{5x\left(x-y\right)}\)

NT

Nguyễn Thùy Chi

18 tháng 12 2021

cho x,y,z là các số dương thoả mãn \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\)=6

Chứng minh \(\dfrac{1}{3x+3y+2z}+\dfrac{1}{3x+2y+3z}+\dfrac{1}{2x+3y+3z}\)≤\(\dfrac{3}{2}\)

#Toán lớp 9

1

NH

Nguyễn Hoàng Minh

18 tháng 12 2021

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

TL

Tuyết Ly

14 tháng 5 2022

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