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Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)
\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)
\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)
\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)
`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`
`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`
`=sqrt{10}-(2+sqrt{10})`
`=-2`
`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`
`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`
`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`
`=(5sqrt5-5sqrt3-4)/2`
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)
Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\frac{7-3}{2}=2\)
Vậy \(P=2\)
\(A=\frac{\sqrt{5-2\sqrt{5}\sqrt{3}+3}}{\sqrt{10}-\sqrt{6}}\)
\(A=\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{10}-\sqrt{6}}\)
\(A=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{10}-\sqrt{6}}\)
\(A=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}\)
\(A=\frac{1}{\sqrt{2}}\)
\(A=\frac{\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}=\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}=\frac{1}{\sqrt{2}}\\ \)