\(ĐKXĐ:x\ge0,x\ne1\)

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

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

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

b/ Ta có: \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

Thay \(x=\left(\sqrt{3}-1\right)^2\) vào (1) ta được:

\(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)= \(\dfrac{\sqrt{3}-1}{4-2\sqrt{3}+\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{4-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}=\dfrac{3\sqrt{3}-1}{13}\)

Vậy giá trị của A khi \(x=4-2\sqrt{3}\) là \(\dfrac{3\sqrt{3}-1}{13}\)

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

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

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

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

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

học tốt nhé anh trai

Cho \(5\sqrt{x}7\) mk viet nham

Sua lai thanh \(5\sqrt{x}-7\)

a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)

=>10 căn x+5-5 chia hết cho 2 căn x+1

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

hay \(x\in\varnothing\)

\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

CHÚC BẠN HỌC TỐT NHÉ

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

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

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

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

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

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

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

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

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

Câu b : Thay \(x=1\dfrac{1}{3}=\dfrac{4}{3}\) vào A ta được :

\(A=\dfrac{2.\dfrac{4}{3}+\sqrt{\dfrac{4}{3}}+1}{\dfrac{4}{3}-\sqrt{\dfrac{4}{3}}}=\dfrac{\dfrac{8}{3}+\dfrac{2\sqrt{3}}{3}+\dfrac{3}{3}}{\dfrac{4}{3}-\dfrac{2\sqrt{3}}{3}}=\dfrac{\dfrac{11+2\sqrt{3}}{3}}{\dfrac{4-2\sqrt{3}}{3}}=\dfrac{11+2\sqrt{3}}{4-2\sqrt{3}}\)

Chúc bạn học tốt

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

a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)

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

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

b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)

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

hay x=0

ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

a) Ta có: \(P=\dfrac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\dfrac{6-2\left(\sqrt{a}-1\right)^2}{a\sqrt{a}-1}+\dfrac{2}{\sqrt{a}-1}\)

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

\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{-2a+4\sqrt{a}+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

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

\(=\dfrac{5a-4\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

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

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

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

\(=\dfrac{5\sqrt{a}+1}{a+\sqrt{a}+1}\)

b) Để P=1 thì \(5\sqrt{a}+1=a+\sqrt{a}+1\)

\(\Leftrightarrow a+\sqrt{a}+1-5\sqrt{a}-1=0\)

\(\Leftrightarrow a-4\sqrt{a}=0\)

\(\Leftrightarrow\sqrt{a}\left(\sqrt{a}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\\\sqrt{a}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\left(nhận\right)\\a=16\left(nhận\right)\end{matrix}\right.\)

Vậy: Để P=1 thì \(a\in\left\{0;16\right\}\)

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a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)

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

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

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

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