rút gọn bt sa=u
\(\frac{\sqrt{39x}}{32}\)+23=90=100
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L
1
KT
12 tháng 8 2018
\(A=\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\frac{6}{\sqrt{3}}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=10-2\sqrt{3}+\frac{6}{\sqrt{3}}\)
\(=\frac{10\sqrt{3}-6+6\sqrt{3}}{\sqrt{3}}\)
\(=\frac{16\sqrt{3}-6}{\sqrt{3}}\)
VA
0
TB
25 tháng 7 2018
ĐKXĐ: \(x\ge0\)
\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
MT
6 tháng 9 2015
\(B=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right)\left(\frac{1}{\sqrt{a}}+1\right)\)
\(=\left(\frac{1+\sqrt{a}}{1-a}-\frac{1-\sqrt{a}}{1-a}\right)\left(\frac{\sqrt{a}}{a}+\frac{a}{a}\right)\)
\(=\frac{1+\sqrt{a}-1+\sqrt{a}}{1-a}.\frac{\sqrt{a}+a}{a}\)
\(=\frac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\frac{\sqrt{a}.\left(1+\sqrt{a}\right)}{a}\)
\(=\frac{2}{1-\sqrt{a}}\)
DT
3
5 tháng 12 2016
\(A^2=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(\Leftrightarrow A^2=\frac{\left(a+1\right)^2+a^2\left(a^2+2a+2\right)}{a^2\left(a+1\right)^2}\)
\(\Leftrightarrow A^2=\frac{\left(a+1\right)^2+2\left(a+1\right)a^2+a^4}{a^2\left(a+1\right)^2}\)
\(\Leftrightarrow A^2=\frac{\left(a+1+a^2\right)^2}{a^2\left(a+1\right)^2}\)
TN
5 tháng 12 2016
\(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\frac{1}{a}\right)^2+\frac{1}{\left(a+1\right)^2}-\frac{2}{a}}\)
\(=\sqrt{\left(\frac{a+1}{a}\right)^2+\frac{1}{\left(a+1\right)^2}-\frac{2\left(a+1\right)}{a}\cdot\frac{1}{a+1}}\)
\(=\sqrt{\left(\frac{a+1}{a}-\frac{1}{a+1}\right)^2}=\left|\frac{1}{a}+1-\frac{1}{a+b}\right|\)