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a) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{2}-2\right)\)
\(=\left|\sqrt{2}+1\right|-\sqrt{2}+2\)
\(=\sqrt{2}+1-\sqrt{2}+2\)
\(=3\)
b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)
\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)
\(=-8\sqrt{6}+2\sqrt{6}\)
\(=-6\sqrt{6}\)
c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)
\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)
\(=\left(\sqrt{5}\right)^2-3^2\)
\(=-4\)
a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)
\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)
\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)
\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)
\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)
\(=3\)
a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)
b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)
c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)
d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)
f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{3\sqrt{5}}{2}\)
\(\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}-3}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{2}\)
\(=\dfrac{1-3}{2}\)
\(=-1\)
`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`
`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`
`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`
`=2sqrt2-8/(2sqrt2)`
`=2sqrt2-sqrt2=sqrt2`
`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`
`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`
`=3-sqrt5-2(sqrt5-1)+3sqrt5`
`=3+3sqrt5-3sqrt5+2=5`
Ta có: \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\cdot\sqrt{4.5}+\dfrac{2}{5}\sqrt{50}\right):\dfrac{4}{15}\sqrt{\dfrac{1}{8}}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{9}{4}\sqrt{2}+2\sqrt{2}\right):\dfrac{\sqrt{2}}{15}\)
\(=0\)
b)
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)
a)
\(\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{9-3\sqrt{6}+5\sqrt{6}-10}{5}=\dfrac{-1+2\sqrt{6}}{5}\)
d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
=căn 5-1/2*căn 5
=1/2*căn 5
e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)
f:=2+căn 3+căn 2-2-căn 3=căn 2
1, -6\(\sqrt{6}\)
2, -4
3, 0
4, 3
6, -2