Question

[_Banhdayyy_]

Mn giúp mik câu i,k,m ạ. Mik camon nhìuu

Answer 1

[An Thy]

i) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}=\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}+\left|\sqrt{2}-2\right|\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}+2-\sqrt{2}=\left|\sqrt{2}+1\right|+2-\sqrt{2}=\sqrt{2}+1+2-\sqrt{2}=3\)

k) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}=\sqrt{\dfrac{8-2\sqrt{15}}{2}}-\sqrt{\dfrac{8+2\sqrt{15}}{2}}+\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}+\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}+\sqrt{6}\)

\(=\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}+\sqrt{6}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\sqrt{6}=\dfrac{-2\sqrt{3}}{\sqrt{2}}+\sqrt{6}=-\sqrt{6}+\sqrt{6}=0\)

m) \(2\sqrt{56}-14\sqrt{\dfrac{2}{7}}+\left(\sqrt{7}-\sqrt{2}\right)\sqrt{7}-\dfrac{8\sqrt{2}}{\sqrt{3}-\sqrt{7}}\)

\(=2\sqrt{4.14}-2\sqrt{49.\dfrac{2}{7}}+7-\sqrt{14}+\dfrac{8\sqrt{2}.\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}\)

\(=4\sqrt{14}-2\sqrt{14}+7-\sqrt{14}+\dfrac{8.\left(\sqrt{14}+\sqrt{6}\right)}{4}\)

\(=\sqrt{14}+7+2\left(\sqrt{14}+\sqrt{6}\right)=7+3\sqrt{14}+2\sqrt{6}\)

Answer 2

[Akai Haruma]

Lời giải:
i.

\(=\sqrt{(\sqrt{2}+1)^2}+|\sqrt{2}-2|=|\sqrt{2}+1|+|\sqrt{2}-2|=\sqrt{2}+1+2-\sqrt{2}=3\)

k.

\(=\frac{1}{\sqrt{2}}(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}+\sqrt{12})\)

\(=\frac{1}{\sqrt{2}}(\sqrt{(\sqrt{3}-\sqrt{5})^2}-\sqrt{(\sqrt{3}+\sqrt{5})^2}+2\sqrt{3})\)

\(=\frac{1}{\sqrt{2}}(|\sqrt{3}-\sqrt{5}|-|\sqrt{3}+\sqrt{5}|+2\sqrt{3})=\frac{1}{\sqrt{2}}(-2\sqrt{3}+2\sqrt{3})=0\)

m.

\(=4\sqrt{14}-2\sqrt{14}+7-\sqrt{14}-\frac{8\sqrt{2}(\sqrt{3}+\sqrt{7})}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})}\)

\(=\sqrt{14}+7-\frac{8(\sqrt{14}+\sqrt{6})}{-4}=\sqrt{14}+\sqrt{7}+2(\sqrt{14}+\sqrt{6})=3\sqrt{14}+\sqrt{7}+2\sqrt{6}\)