a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:

\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)

b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)

\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)

\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:

\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)

\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

Bài mình làm sai à hay sao mà t]ick sai thế???

a. 2x(x + y) - y(y + 2x) = 2x² + 2xy - y² - 2xy = 2x² - y²

b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

Phần c nản quá.

a) 2x(x + y) - y(y + 2x) 

= 2x² + 2xy - y² - 2xy

= 2x² - y²

b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)

= \(\frac{x^3-4x^2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}\)

= \(\frac{x^3-4x^2+2x-2+x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{x^3-3x^2+3x-1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{\left(x-1\right)^3}{\left(x^2+x+1\right)\left(x-1\right)}\)

\(=\frac{\left(x-1\right)^2}{x^2+x+1}\)

a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)

\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)

\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)

b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)

\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)

\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)

a) A= \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\)

\(ĐK:3x^2-7x+2\ne0\)

\(\Leftrightarrow\orbr{\begin{cases}x\ne\frac{1}{3}\\x\ne2\end{cases}\left(^∗\right)}\)

=> 3x² + 5x + 2 =0

<=> 3x² + 3x + 2x +2 = 0

<=> 3x .( x + 1 ) + 2 .( x + 1 ) =0

<=> (  x + 1 )(3x + 2 ) =0

<=> \(\orbr{\begin{cases}x+1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\left(t/m\left(^∗\right)\right)\end{cases}}}\)

Vậy x = -2/3 

b) \(B=\frac{2x^2+10x+12}{x^3-4x}=0\left(ĐK:x\ne0;x^2\ne4\Leftrightarrow x\ne0;x\ne\pm2\right)\)

<=> 2x²+ 10x + 12 = 0

<=> x² + 5x+ 6 =0

<=> ( x + 2 ) ( x + 3 ) =0\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=-3\left(t/m\right)\end{cases}}\) 

Vậy x = -3 

c)\(C=\frac{x^3+x^2-x-1}{x^3+2x-5}=0\)                         \(ĐK:x^3+2x-5\ne0\left(^∗\right)\)

<=> x³ + x² -x -1 =0

<=> ( x - 1 )(x² + 2x + 1 ) 

<=> ( x-1 ) (x+1)² = 0

<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(t/m\left(^∗\right)\right)\\x=-1\left(t/m\left(^∗\right)\right)\end{cases}}}\)

Vậy x = { 1 ; -1 }

a) A = \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\) (ĐKXĐ: x khác 1/3, x khác 2)

<=> 3x^2 + 5x - 2 = 0

<=> (3x - 1)(x + 2) = 0

<=> 3x - 1 = 0 hoặc x + 2 = 0

<=> 3x = 1 hoặc x = -2

<=> x = 1/3 (ktm) hoặc x = -2 (tm)

=> x = -2

b) B = \(\frac{2x^2+10x+12}{x^3-4x}=0\) (ĐKXĐ: x khác 0, x khác +-2)

<=> \(\frac{2\left(x^2+5x+6\right)}{x\left(x^2-4\right)}=0\)

<=> \(\frac{2\left(x+2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{2\left(x+3\right)}{x\left(x-2\right)}=0\)

<=> 2(x + 3) = 0

<=> x + 3 = 0

<=> x = -3

c) C = \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) (ĐKXĐ: x khác x^3 + 2x - 5)

<=> \(\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)

<=> \(\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}=0\)

<=> \(\frac{\left(x+1\right)\left(x-1\right)\left(x+1\right)}{x^3+2x-5}=0\)

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

<=> x + 1 = 0 hoặc x - 1 = 0

<=> x = -1 hoặc x = 1

cảm ơn