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13 tháng 8 2019

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

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

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

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

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

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

=\(\frac{2}{x+\sqrt{x}+1}\)

P>0 dựa vào dkxd

13 tháng 8 2019

b giống a

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:  \(a\ge0;a\ne1\)

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

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

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

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

8 tháng 8 2018

Bài 1:

a, (Xin được sửa đề bài) \(C=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}\)

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

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

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

b, \(D=\sqrt{m^2}-\sqrt{m^2-10m+25}\)

\(=m-\sqrt{\left(m-5\right)^2}\)

\(=m-m+5=5\)

Bài 2:

a, \(VT=\sqrt{x+2\sqrt{x-2}-1}.\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)\)

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

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

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

\(=\left(x-3\right):\left(\sqrt{x}-\sqrt{3}\right)\)

\(=\sqrt{x}+\sqrt{3}=VP\)

b, \(VT=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a+1-2\sqrt{a}}\)

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

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

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

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

\(=\frac{\sqrt{a}-1}{\sqrt{a}}=VP\)

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

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

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

3 tháng 4 2017

mình giải thế này

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

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

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

b)\(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow\sqrt{x}-1< 0\)

\(\Rightarrow-x\left(\sqrt{x}-1\right)>0\)vì \(x>0\)

xong rồi nhé :)

3 tháng 4 2017

Hình như kết quả rút gọn là  \(\sqrt{x}-x\)