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a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
a)\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
b) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}=4\sqrt{3}\)
c)\(\sqrt{12}+2\sqrt{75}-3\sqrt{48}-\frac{2}{7}\sqrt{147}=2\sqrt{3}+10\sqrt{3}-12\sqrt{3}-2\sqrt{3}=-2\sqrt{3}\)
d) \(\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{9-4\sqrt{5}}\)
\(=\left|3+\sqrt{5}\right|-\sqrt{\left(\sqrt{5}-2\right)^2}=3+\sqrt{5}-\left|\sqrt{5}-2\right|=3+\sqrt{5}-\sqrt{5}+2=5\)
e) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{\sqrt{5}+\sqrt{2}}{3}\)
\(=\left[\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}=-3\)
Nản k lm nữa ^^
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)
\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)
\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)
\(=6+\sqrt{5}\)
\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)
\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)
\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}-2+\sqrt{5}\)
\(=-3+\sqrt{3}+\sqrt{5}\)
\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{3}\)
\(=\sqrt{3}-1+2\sqrt{3}\)
\(=-1+3\sqrt{3}\)
\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\sqrt{6}\)
\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)
Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)
Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
.... Đúng thì ủng hộ nha ....
Kết bạn với mình ... ;) ;)