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30 tháng 10 2020

1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

\(=2+\sqrt{3}-2+\sqrt{3}=VP\)

30 tháng 10 2020

Bài 1.

Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)

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

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

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

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

17 tháng 8 2015

1)))))))

\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

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

 

17 tháng 8 2015

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

\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)

27 tháng 4 2019

\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

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

         \(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)

        \(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)

       \(=-3\)

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

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

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

    \(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, Ta có \(B< A\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)

\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)

\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)

Vậy ...

21 tháng 4 2020

@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm

còn câu b kl sai r nha

21 tháng 4 2020
https://i.imgur.com/K1Kg6qE.jpg
21 tháng 9 2018

DKXD: \(x\ge0;x\ne1\)

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

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

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

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

b) \(\sqrt{x}^3+1>0;\left(x+1\right)^2>0\) mọi x

\(\Rightarrow A>0\forall x\ge0;x\ne1\)