\(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(2a+2\sqrt{ab}+2b\right)}
\)
a. Rút gọn P
b. Tìm giá trị nguyên của a để giá trị P nguyên
K
Khách
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26 tháng 12 2015
\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)
\(=\left(\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{3a}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\left(a+\sqrt{ab}+b\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{2\left(a+\sqrt{ab}+b\right)}{\cdot\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\left(a-2\sqrt{ab}+b\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{2}{a-1}\)
T
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\(\ge0\)
HA
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!!
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 }