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

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

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

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

2 tháng 9 2021

giải chi tiết hơn được không ạ

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

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

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

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

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

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

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

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

29 tháng 5 2022

làm chi tiết hộ e vs ạ

NV
22 tháng 7 2021

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

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

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

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

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 \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

18 tháng 10 2023

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

19 tháng 1 2022

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

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

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

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

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

6 tháng 3 2022

Câu 1 : 

a, \(=8+4-2.6=12-12=0\)

b, đk : x > 0 ; x khác 1 

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

6 tháng 9 2021

a, ĐK: \(x\ge0;x\ne1\)

\(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right).\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\left[\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

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

6 tháng 9 2021

b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}\in Z\)

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

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ta có điều phải chứng minh.