Rút gọn biểu thức sau:
\(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
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12 tháng 6 2023
\(\sqrt{9-3\sqrt{8}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}+\sqrt{5-2\sqrt{6}}-\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{6}+\sqrt{2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{6}-\sqrt{3}\right|-\sqrt{6}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}+\sqrt{3}-\sqrt{2}\) (do \(\sqrt{6}-\sqrt{3}>0;\sqrt{3}-\sqrt{2}>0\))
\(=0\)
12 tháng 6 2023
\(=\sqrt{9-6\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{3}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}\)
\(=0\)
25 tháng 8 2021
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
AT
7 tháng 7 2021
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
TG
7 tháng 7 2021
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
TL
1
16 tháng 9 2021
\(A=\sqrt{5-2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}\)
=0
4 tháng 7 2023
\(=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{5}+1}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{5}+1}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}\)
\(=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{4}\)
\(=\sqrt{2}\cdot\dfrac{2}{4}=\dfrac{1}{\sqrt{2}}\)
2 tháng 10 2021
\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)
\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)
\(=4\)
24 tháng 11 2021
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
2 tháng 10 2021
a: Ta có: \(A=\left(\sqrt{48}-2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-2\sqrt{45}-\sqrt{3}\)
\(=\left(2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-6\sqrt{5}-\sqrt{3}\)
\(=2\sqrt{15}+10-6\sqrt{5}-\sqrt{3}\)
b: Ta có: \(B=\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5}+\sqrt{2}}\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{2}}{9+6\sqrt{2}}=\dfrac{-8+6\sqrt{2}}{3}\)
Đặt \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}.\)
\(\Rightarrow x^3=\sqrt{5}+2-3\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\left(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\right)-\sqrt{5}+2\)
\(=4-3\sqrt[3]{5-4}.x\)( Vì \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\))
\(=4-3x\)
\(\Rightarrow x^3+3x-4=0\Leftrightarrow\left(x^3-1\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+4\right)=0\Leftrightarrow x-1=0\)( Vì \(x^2+x+4=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
\(\Leftrightarrow x=1\)hay \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1\)