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

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

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

làm chi tiết hộ e vs ạ

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

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

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

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

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

Câu 1 : 

a, \(=8+4-2.6=12-12=0\)

b, đk : x > 0 ; x khác 1 

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

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

a, x > 0 ; x khác 1 

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

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

b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)

\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm) 

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

b: Để P=1 thì \(x-\sqrt{x}-2=0\)

hay x=4

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

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

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

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

a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)

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

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

\(\Leftrightarrow x=1\left(loại\right)\)

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

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

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

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

b. Đặt \(B=A-2x\)

\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)

\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

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

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

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

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

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

b/

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

Để B nguyên

\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)

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

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

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

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

\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)

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

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

=1-x

b) Để A=-1 thì 1-x=-1

hay x=2