ĐK : \(x\ne2\); \(x\ne-2\)

a) \(A=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)

\(=\frac{x^3-x.\left(x+2\right)-2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-2x-2x+4}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)

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

b)  -  Để A > 0 thì   x - 1 > 0  =>  x > 1

     -  Để A < 0 thì   x - 1 < 0  =>  x < 1

c) Để  | A | = 5 thì   | x-1 | = 5

+ Nếu \(x-1\ge0\) thì \(x\ge1\) , ta có phương trình

x - 1 = 5 => x = 6 ( thỏa mãn ) 

+ Nếu x - 1 < 0 thì x < 1 , ta có phương trình : 

-x + 1 = 5  < = >  -x = 4  <=>  x = -4  ( thỏa mãn )

Vậy tập nghiệm của phương trình là S = { -4 ; 6 }

a) `(x+y)^2+(x-y)^2=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2`

b) `(a-b^2)(a+b^2)=a^2-(b^2)^2=a^2-b^4`

ĐK:  \(x>0;x\ne1\)

\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)

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

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

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

\(A>-1\) \(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}+1>0\)  \(\Leftrightarrow\)\(\frac{2\sqrt{x}-1}{\sqrt{x}}>0\)

Do  \(\sqrt{x}>0\)  \(\Rightarrow\)\(2\sqrt{x}-1>0\)\(\Leftrightarrow\)\(2\sqrt{x}>1\)\(\Leftrightarrow\)\(\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow\)\(x>\frac{1}{4}\)

Vậy  \(x>\frac{1}{4}\)\(\left(x\ne1\right)\)thì  A > - 1

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

Ta có: \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2-2\sqrt{x}+1}\)

\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

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

Để \(A>-1\)thì \(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)\(\Leftrightarrow\sqrt{x}-1>-\sqrt{x}\)\(\Leftrightarrow2\sqrt{x}>1\)

\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)

Vậy \(A>-1\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)

ĐKXĐ : x > 0

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)