rút gọn các biểu thức sau
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)với \(a\ge0;b\ge0;a\ne b\)
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Những câu hỏi liên quan
3 tháng 2 2022
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
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\(\ge0\)
TH
4 tháng 10 2015
\(=\left(\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}\right).\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}\)
\(=\left(\frac{4\sqrt{ab}+\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right).\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\frac{4\sqrt{ab}+a-2\sqrt{ab}+b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right).\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=1\)
tick cho mình nha
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{4+4\sqrt{3}+3}\)
\(=2-\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}=2-\sqrt{3}+\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(a-2\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)
\(=\left|2-\sqrt{3}\right|+\sqrt{3+4\sqrt{3}+4}\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=2-\sqrt{3}+\left|\sqrt{3}+2\right|\)
\(=2-\sqrt{3}+\sqrt{3}+2\)
\(=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)( \(\hept{\begin{cases}a,b\ge0\\a\ne b\end{cases}}\))
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-2\sqrt{ab}+b\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-2\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-b}{a-b}=1\)