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20 tháng 8 2017

\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)

\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)

\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)

\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)


Điều kiện : a, b\(\ge0\)

11 tháng 8 2018

\(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\div\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)

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

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

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

11 tháng 8 2018

giải tiếp

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

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

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

7 tháng 12 2016

mi tích tau tau tích mi xong tau trả lời nka

 việt nam nói là làm

10 tháng 8 2019

Em kiểm tra lại đề bài nhé!

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

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

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

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

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

29 tháng 8 2018

Ta có:

\(B=\frac{\frac{\left(a-b\right)^3}{\left(\sqrt{a}+\sqrt{b}\right)^3}+2a\sqrt{a}+b\sqrt{b}}{a\sqrt{a}+b\sqrt{b}}+\frac{3\left(\sqrt{ab}-b\right)}{a-b}\)

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

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

\(=\frac{3a\sqrt{a}-3a\sqrt{b}+3\sqrt{a}b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}+\frac{3\left(\sqrt{ab}-b\right)}{a-b}\)

\(=\frac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}+\frac{3\left(\sqrt{ab}-b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{3\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{3\left(\sqrt{ab}-b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+3\left(\sqrt{ab}-b\right)}{a-b}\)

\(=\frac{3a-3b}{a-b}\)

\(=3\)

=.= hok tốt!!