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12 tháng 8 2019

a) đk : \(x\ge0\) ; \(x\ne1\)

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

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

b) đk : \(x\ne0;x\ne1\)

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

11 tháng 7 2018

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

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

\(=1-a+a=1\)

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

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

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

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

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

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

16 tháng 5 2021

a, Với \(x\ge0;x\ne1\)

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

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

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

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

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

16 tháng 5 2021

Bạn ghi chuẩn đề chưa vậy

26 tháng 3 2019

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

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

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

25 tháng 10 2020

Bài làm

Rút gọn

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

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

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

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

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

Tính:

\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)

\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)

\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)

\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)

\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)

\(=6-7\sqrt{3}+7\sqrt{3}\)

\(=6\)

25 tháng 10 2020

Bài làm

\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)

\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)

\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)

\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)

\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)

\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)

\(=5-4\)

\(=1\)

7 tháng 7 2018

mk làm luôn

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)

=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)

=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)

=