Rút gọn
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{18}}\)
\(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right).\left(15+2\sqrt{6}\right)\)
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1 tháng 10 2021
a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
24 tháng 9 2021
\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)
24 tháng 9 2021
a) \(=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}\)
b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)
c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)
12 tháng 12 2018
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
18 tháng 9 2019
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
19 tháng 9 2019
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
b) Ta có: \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(\frac{5+2\sqrt{6}+2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\frac{5+2\sqrt{6}+10-4\sqrt{6}}{25-24}\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=15^2-\left(2\sqrt{6}\right)^2\)
\(=225-24=201\)