a, Với \(x=3\)\(=>A=\frac{x-1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)

Vậy A = 1 khi x = 3

b, Ta có : \(B=\frac{1}{x}-\frac{x}{2x+1}+\frac{2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

Ta có : \(A=\frac{x-1}{2};B=\frac{x-1}{2x+1}\)

\(=>C=A:B=\frac{x-1}{2}:\frac{x-1}{2x+1}=\frac{2x+1}{2}=x+\frac{1}{2}\)

đề sai bạn ơi 

Đề sai rồi bạn

a) đặt mẫu chứng là x-2

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

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

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

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

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

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

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

b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)

=>\(x^2+x+1-x⋮x^2+x+1\)

=>\(x⋮x^2+x+1\)

=>\(x^2+x⋮x^2+x+1\)

=>\(x^2+x+1-1⋮x^2+x+1\)

=>\(-1⋮x^2+x+1\)

=>\(x^2+x+1\in\left\{1;-1\right\}\)

=>\(x^2+x+1=1\)

=>x²+x=0

=>x(x+1)=0

=>\(x\in\left\{0;-1\right\}\)

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

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

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

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

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

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

b) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(P=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\), ta được:

\(P=\left[\left(\dfrac{1}{4}\right)^2+2\cdot\dfrac{1}{4}+1\right]:\left[\left(2\cdot\dfrac{1}{4}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\left(\dfrac{1}{16}+\dfrac{1}{2}+1\right):\left[\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\dfrac{25}{16}:\dfrac{51}{32}=\dfrac{25}{16}\cdot\dfrac{32}{51}=\dfrac{50}{51}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(P=\dfrac{50}{51}\)

1: Ta có: \(A=\left(\dfrac{x^2-16}{x-4}-1\right):\left(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}+\dfrac{x+2-x^2}{x^2-2x-3}\right)\)

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

\(=\left(x+3\right):\dfrac{x^2+x-2x-2+x^2-9-x^2+x+2}{\left(x-3\right)\left(x+1\right)}\)

\(=\left(x+3\right):\dfrac{x^2-9}{\left(x-3\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+3\right)\left(x-3\right)\left(x+1\right)}{x^2-9}\)

\(=x+1\)

ĐKXĐ: \(x\notin\left\{4;3;-1\right\}\)

2: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì \(x+1⋮x^2+x+1\)

\(\Leftrightarrow x^2+x⋮x^2+x+1\)

\(\Leftrightarrow x^2+x+1-1⋮x^2+x+1\)

mà \(x^2+x+1⋮x^2+x+1\)

nên \(-1⋮x^2+x+1\)

\(\Leftrightarrow x^2+x+1\inƯ\left(-1\right)\)

\(\Leftrightarrow x^2+x+1\in\left\{1;-1\right\}\)

\(\Leftrightarrow x^2+x\in\left\{0;-2\right\}\)

\(\Leftrightarrow x^2+x=0\)(Vì \(x^2+x>-2\forall x\))

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

Vậy: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì x=0

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

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

b: Khi x=9 thì P=9-3+1=7

c: P=3

=>x-căn x-2=0

=>(căn x-2)(căn x+1)=0

=>x=4

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

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

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

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

\(=\left(\sqrt{2}-1\right)^2\)

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

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

\(=\dfrac{1}{\sqrt{2}-1}\)

\(=\sqrt{2}+1\)

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)