1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

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

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

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

Bạn vào biểu tượng \(\Sigma\) để nhập biểu thức cho chính xác nhé

\(1)\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)\left(x-1\right)\\ =x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x-1\right)^2\\ =6x^2+2-6\cdot\left(x^2-2x+1\right)\\ =6x^2+2-6x^2+12x-6\\ =12x-4\)

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

\(3)\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-4\right)\left(x+4\right)\\ =x^3-3x^2+3x-1-(x^3+8)+3\cdot\left(x^2-16\right)\\ =x^3-3x^2+3x-1-x^3-8+3x^2-48\\ =3x-55\)

Thanks bạn

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

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

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

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

\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)

\(\Leftrightarrow6x=2x^2+4\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)

\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)

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