\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\left(x\ne1;x\ne3\right)\)

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

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

\(\Rightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

<=> x=0 hoặc x=1

Vậy x=0; x=1

\(ĐKXĐ:x\ne3;x\ne1\)

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

\(\Leftrightarrow x^2-1-x+3=2\)

\(\Leftrightarrow x^2-x=0\Leftrightarrow x=0\)(vì x khác 1)

Vậy x = 0

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)

\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)

=> PT đúng với mọi x khác \(\pm5\)

Refund QB nhìn logic :V 

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)

\(20x=2x^2+50\)

\(20x-2x^2-50=0\)

\(2\left(10x-x^2-25\right)=0\)

\(-x^2+10x+25=0\)

\(x^2-10x+25=0\)

\(x^2-2\left(x\right)\left(5\right)+5^2=0\)

\(\left(x-5\right)^2=0\)

\(x-5=0\Leftrightarrow x=5\)

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

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

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

\(2+\frac{1}{x}=x+\frac{1}{x}\)

\(2=x\Leftrightarrow x=2\)

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

\(\Leftrightarrow\frac{2x+1}{x}=\frac{x^2+1}{x}\)

\(\Rightarrow\orbr{\begin{cases}2x+1=x^2+1=0\\2x+1=x^2+1\end{cases}}\)

Mà \(x^2+1>0\)nên \(x^2-2x=0\Leftrightarrow x=2\)(vì x khác 0)

Vậy x = 2

1/x 1/x và x ngũ 2

2 + 1x = (1x + 2)(x² + 1)

<=> 2 + x = 2x² + 2 + x³ + x

<=> x = 2x² + x³ + x

<=> 0 = 2x² + x³

<=> 2x² + x³ = 0

<=> x²(2 + x) = 0

<=> x = 0 hoặc 2 + x = 0

<=> x = 0 hoặc x = -2

\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)

\(x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(2x^2-2x=x+3-x^2-3x\)

\(2x^2-2x=-2x+3-x^2\)

\(2x^2=3-x^2\)

\(2x^2+x^2=3\)

\(3x^2=3\Leftrightarrow x^2=1\Leftrightarrow x=\pm\sqrt{1}\)

tớ n g u nên cần tg suy nghĩ thêm :v 

câu a tìm ra r nè , vất vả :v ( kiên trì lắm đấy )

\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2+1\right)\)

\(9x^3+9x^2-4x-4-3x^2-3x-2x^2-2=0\)

\(6x^3+7x^2-7x-6=0\)

\(\left(6x^2+13x+6\right)\left(x-1\right)=0\)

\(Th1:6x^2+9x+4x+6=0\)

\(\Leftrightarrow\left[3x\left(2x+3\right)+2\left(2x+3\right)\right]=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\3x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}}\)

\(Th2:x-1=0\Leftrightarrow x=1\)

4x² + 8x + x +4 = 0

\(\Leftrightarrow\)4x² + 9x +4 = 0

\(\Leftrightarrow\)[(2x)² + 2.2x. 9/4 + 81/16]+4 - 81/16=0

\(\Leftrightarrow\) (2x+9/4)²=17/16

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+\frac{9}{4}=\frac{\sqrt{17}}{4}\\2x+\frac{9}{4}=\frac{-\sqrt{17}}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{17}-9}{8}\\x=\frac{-\sqrt{17}-9}{8}\end{cases}}\)

\(4+2x\left(2x+4\right)=-x\)

\(\Leftrightarrow4x^2+8x+x+4=0\)

\(\Leftrightarrow4x^2+9x+4=0\)

Ta có \(\Delta=9^2-4.4.4=17,\sqrt{\Delta}=\sqrt{17}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-9+\sqrt{17}}{8}\\x-\frac{-9-\sqrt{17}}{8}\end{cases}}\)

x.(x+2)+2=-x

\(\Leftrightarrow\)x² + 2x + x + 2=0

\(\Leftrightarrow\)x² + 3x + 2=0

\(\Leftrightarrow\)(x+1).(x+2) =0

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}}\)

\(x\left(x+2\right)+2=-x\)

\(x^2+2x+2=-x\)

\(x^2+2x+2+x=0\)

\(x^2+3x+2=0\)

\(\left(x+1\right)\left(x+2\right)=0\)

\(Th1:x+1=0\Leftrightarrow x=-1\)

\(Th2:x+2=0\Leftrightarrow x=-2\)