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3 tháng 8 2017

sửa lại đề đi \(\sqrt{a+\sqrt{b}}\) hay căn a+căn b

3 tháng 8 2017

đk \(a>0;b>0;a\ne b\)\(R=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{a}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)-\frac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}\)

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

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

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

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

\(R=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{a}+\sqrt{b}}{2}=\frac{-2\sqrt{b}}{2}=-\sqrt{b}\)

b) \(R=-1\Leftrightarrow-1=-\sqrt{b}\Leftrightarrow1=\sqrt{b}\Leftrightarrow b=1\)

b=(a+1)2 <=> 1=(a+1)2 <=> a+1=1 <=> a=0

vậy a = 0 ; b=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 } 

19 tháng 8 2019

\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}.\)

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

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

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

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

\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

19 tháng 8 2019

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

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

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

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

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

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\(\ge0\)