` @ \color{Red}{m}`

` \color{lightblue}{Answer}`  

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

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\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)

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

\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)

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

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

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

\(=\dfrac{x}{x+1}\)

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\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)

\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)

\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)

\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)

\(=\dfrac{x+6}{2x^2+6x}\)

==========================

\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)

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

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

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

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

\(=\dfrac{1}{x+1}\)

a)

\(DKXD:\left[{}\begin{matrix}x^2+x\ne0\\x\ne0\\x+1\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

b)

\(\left(\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)

\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)

\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)

\(=\left(\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)

\(=\dfrac{x^2+3x}{x\left(x+1\right)}\cdot\dfrac{x+1}{3}\\ =\dfrac{x\left(x+3\right)\cdot\left(x+1\right)}{x\left(x+1\right)\cdot3}\\ =\dfrac{x+3}{3}\)

ĐKXĐ: \(x\notin\left\{1;\dfrac{1}{2}\right\}\)

\(\left(\dfrac{1}{x-1}+2+\dfrac{2x^3+x^2-x}{1-x^3}\right):\dfrac{1-2x}{x^3+x-2}\)

\(=\left(\dfrac{1}{x-1}+2-\dfrac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^3+x-2}{1-2x}\)

\(=\dfrac{x^2+x+1+2\left(x^3-1\right)-2x^3-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^3-x^2+x^2-x+2x-2}{-\left(2x-1\right)}\)

\(=\dfrac{2x+1+2x^3-2-2x^3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+2\right)}{-\left(2x-1\right)}\)

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

Bạn ghi lại đề đi bạn. Khó hiểu quá!

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

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

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

\(=x-1\)

b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)

\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{-1}{y}\)

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

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

\(=\dfrac{x^2+x-2+x^2-x+2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)

\(=\dfrac{2x^2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)

\(=\dfrac{2x\left(x^2-2\right)\left(x-3\right)}{2x\left(x-2\right)\cdot\left(x+2\right)^2}=\dfrac{\left(x^2-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)^2}\)

a, đề này chắc sai ở đoạn \(\dfrac{2x}{x^2-3}\) sửa thành \(\dfrac{2x}{x-3}\)

\(=>đk:x\ne1,x\ne3\)

\(=>A=\dfrac{2x}{x-3}+\dfrac{2x}{x^2-4x+3}+\dfrac{x}{x-1}\)

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

\(=\dfrac{3x^2-3x}{\left(x-1\right)\left(x-3\right)}=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)

b, \(A=\dfrac{3x}{x-3}=3+\dfrac{9}{x-3}\)

A nguyên <=>\(x-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

\(=>x\in\left\{4;2;6;0;12;-6\right\}\left(TM\right)\)

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

a: \(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2-1}{2x+1}\)

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

b: Để \(P=\dfrac{3}{x-1}\) thì \(\dfrac{3}{x-1}=\dfrac{2}{2x+1}\)

=>6x+3=2x-2

=>4x=-5

hay x=-5/4