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

b, \(P=2022\)

\(\Leftrightarrow\dfrac{4}{x+1}=2022\\ \Leftrightarrow4=2022x+2022\\ \Leftrightarrow2022x=-2018\\ \Leftrightarrow x=-\dfrac{1009}{1011}\)

c, P nguyên 

\(\Leftrightarrow\dfrac{4}{x+1}\in Z\\ \Rightarrow4⋮\left(x+1\right)\\ \Rightarrow x+1\inƯ\left(4\right)\)

Ta có bảng:

Vậy \(x\in\left\{-5;-3;-2;0;3\right\}\)

a. \(A=\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{4x+2}{x^2-1}\)

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

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

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

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

b) Ta có: \(A=\dfrac{4}{x-1}=\dfrac{4}{2015}\) (ĐK: \(x\ne\pm1\) )

\(\Leftrightarrow8060=4\left(x-1\right)\)

\(\Leftrightarrow8060=4x-4\)

\(\Leftrightarrow8064=4x\)

\(\Leftrightarrow x=\dfrac{8064}{4}=2016\left(tm\right)\)

c) Ta có: \(\dfrac{4}{x-1}\left(x\ne1\right)\)

Để \(\dfrac{4}{x-1}\) nhận giá trị nguyên thì \(4:\left(x-1\right)\Leftrightarrow x-1\in\text{Ư}\left(4\right)=\left\{1;4;2\right\}\)

Vậy với x ∈ {2; 5; 3; 0; -1; -3} thì biểu thức \(\dfrac{4}{x-1}\) nhận giá trị nguyên

d) Thay \(x=-\dfrac{1}{2}\) vào biểu thức A ta được:

\(\dfrac{4}{-\dfrac{1}{2}-1}=-3\)

Vậy biểu thức A có giá trị -3 tại \(x=-\dfrac{1}{2}\)

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

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

b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)

Khi x=-1/2 thì B=2/5

c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

a, đk : x khác -2 ; 2 

\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

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

b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)

Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)

c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)

a, điều kiện xác định là \(x\ne1;x\ne-1\)

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

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

\(=\frac{3}{x-1}\)

b, để \(\frac{3x+3}{x^2-1}=-2\Rightarrow\frac{3}{x-1}=-2\)

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

\(\Rightarrow-2x=1\)

\(\Rightarrow x=-\frac{1}{2}\)

a. ĐKXĐ: x² - 1\(\ne\)0 (=) x \(\ne\)\(\pm\)1

b. \(\frac{3x+3}{x^2-1}\)

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

\(=\frac{3}{x+1}\)với x \(\pm\)1

c. \(\frac{3}{x+1}=-2\)

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

\(-2x-2=3\)

\(-2x=5\)

\(x=-\frac{5}{2}\)(t/m đk)

Aww... Sr

Câu b và c phải như này cơ:

b. \(\frac{3x+3}{x^2-1}\)

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

\(=\frac{3}{x-1}\)với x \(\ne\pm1\)

c. \(\frac{3}{x-1}=-2\Leftrightarrow\left(x-1\right).\left(-2\right)=3\)

\(-2x+2=3\)

\(-2x=1\)

\(x=-\frac{1}{2}\)với x \(\ne\pm1\)

a: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

a: \(P=\dfrac{x^2+x-x^2+x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)