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

ĐKXĐ: \(x\ne1;\) \(x\ne-1\)

Quy đồng mẫu hai vế và khử mẫu:

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

Suy ra: ( x + 1 )² - ( x - 1 )² = 4

<=> x² + 2x + 1 - (x² - 2x + 1) = 4

<=> x² + 2x + 1 - x² + 2x - 1 = 4

<=> 4x = 4

<=> x = 4 : 4

<=> x = 1 (Không thỏa ĐKXĐ)

Vậy: phương trình vô nghiệm S=\(\varnothing\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\) ĐKXĐ : \(x\ne\pm1;\pm2\)

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

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

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

\(x^2+2x+1-x^2+2x-1=4\)

\(4x=4\Leftrightarrow x=1\)Theo ĐKXĐ : ko thỏa mãn => vô nghiệm 

\(\frac{5x-2}{3}=\frac{5-3x}{2}\)

<=> \(\frac{\left(5x-2\right).2}{3.2}=\frac{\left(5-3x\right).3}{2.3}\)

<=> \(\frac{10x-4}{6}=\frac{15-9x}{6}\)

Suy ra: 10x - 4 = 15 - 9x

<=> 10x + 9x = 15 + 4

<=> 19x = 19

<=> x = 19:19

<=> x = 1

Vậy: S={1}

Trả lời :

\(\frac{5x-2}{3}=\frac{5-3x}{2}\)

\(\Rightarrow2\times\left(5x-2\right)=3\times\left(5-3x\right)\)

\(\Rightarrow10x-4=15-9x\)

\(\Rightarrow10x+9x=15+4\)

\(\Rightarrow19x=19\)

\(\Rightarrow x=1\)

(6+x)(x-3)=2(x-3)

(6+x)(x-3)-2(x-3)=0

(x-3)(6+x-2)=0

(x-3)(x+4)=0

\(\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}}\)

Vậy x=3,x=-4