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16 tháng 11 2020

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

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

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

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

17 tháng 6 2020

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

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

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

=22-(\(\sqrt{3}\))2=4-3=1

B=(\(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\)) . (\(a\sqrt{b}-b\sqrt{a}\))

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

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

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

=b-a

Ta có: \(A=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\cdot\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)

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

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

\(=\frac{3\sqrt{3}+5}{\sqrt{3}+1}\cdot\frac{3\sqrt{3}-5}{\sqrt{3}-1}\)

\(=\frac{2}{2}=1\)

24 tháng 9 2020

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

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)

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

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

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

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

30 tháng 7 2019

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

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

\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

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

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

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

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

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

Cho e xin cảm ơn trc ak

NV
8 tháng 10 2019

ĐKXĐ:...

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

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

Câu sau đề đúng ko bạn? Thế này thì ko rút gọn được, tử số là \(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\) thì mới rút được

11 tháng 10 2019

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