\(A=\frac{-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-...">
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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!!

27 tháng 9 2016

dài thế

27 tháng 9 2016

\(Ờ,\)\(dài\)\(thật\)

AH
Akai Haruma
Giáo viên
22 tháng 10 2020

Lời giải:

ĐK: $a, b>0$
\(A=\frac{\sqrt{ab}(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}:\frac{2}{\sqrt{a}+\sqrt{b}}=(\sqrt{a}-\sqrt{b}).\frac{\sqrt{a}+\sqrt{b}}{2}=\frac{a-b}{2}\)

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

3 tháng 4 2020

a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)

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

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

          = \(\frac{2}{a-1}\)

b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1 

=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 } 

26 tháng 12 2015

\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)

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

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

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

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

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

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

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

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

15 tháng 6 2017

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

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

\(=a-b\)

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

\(=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\left[\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right]\)

\(=\frac{a-b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\left[\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right]\)

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