Rút gọn biểu thức: \(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\)
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VL
2
H
5 tháng 4 2021
\(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}=\sqrt{a}-\left(\sqrt{a}+2\right)=-2\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
5 tháng 4 2021
Ta có: \(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}-\sqrt{a}-2=-2\)
11 tháng 8 2015
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{6-2+2\sqrt{3}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=1+\sqrt{3}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
23 tháng 11 2025
Ta có: \(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)}{\sqrt{a}-2}+\frac{\sqrt{a}-2}{\sqrt{a}+2}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)^2+\left(\sqrt{a}-2\right)^2-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{-a-4\sqrt{a}-4+a-4\sqrt{a}+4-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{-4a-8\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=-\frac{4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=-\frac{4\sqrt{a}}{\sqrt{a}-2}\)
Ta có: \(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\)
\(=\frac{-2}{\sqrt{a}-2}+\frac{\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-2\sqrt{a}+\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
Ta có: \(\left(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\right):\left(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\right)\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}\cdot\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+3}=\frac{4a}{\sqrt{a}-3}\)
7 tháng 12 2018
\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}=\) \(\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)
\(=\frac{x+6\sqrt{x}+9}{3+\sqrt{x}}\)
\(=\frac{\left(3+\sqrt{x}\right)^2}{3+\sqrt{x}}\)
\(=3+\sqrt{x}\)
7 tháng 12 2018
\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}\left(x\ge0\right)=\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)
\(=\frac{x+\sqrt{6}+9}{3+\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)^2}{3+\sqrt{x}}=3+\sqrt{x}\left(x\ge0\right)\)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}=\sqrt{225-30\sqrt{2}+2}+\sqrt{121+22\sqrt{2}+2}=\sqrt{15^2-15.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{11^2+11.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}=15-\sqrt{2}+11+\sqrt{2}\left(do:15-\sqrt{2}>0;11+\sqrt{2}>0\right)=26\)