Question

1, Rút gọn

A=\( \sqrt{8+2\sqrt15}\)+\(\sqrt{8-2\sqrt15} \)-2\(\sqrt{6-2\sqrt5}\)

B=\(\sqrt{6+2\sqrt5-\sqrt13+\sqrt48}\)

2, Chứng minh

(\(\dfrac{3\sqrt2}{\sqrt27-3}\)-\(\dfrac{\sqrt150}{3}\))*\(\dfrac{1}{\sqrt6}\)=\(\frac{-4}{3}\)

Giúp mik vs. Mai mik phải nộp rồi

Answer 1

Bài 1:

Ta có: \(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-2\cdot\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-2\cdot\left|\sqrt{5}-1\right|\)

\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)

\(=2\sqrt{5}-2\sqrt{5}+2\)

=2

Vậy: A=2

Bài 2: Sửa đề: Chứng minh \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}=\frac{-7+\sqrt{3}}{6}\)

Ta có: \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{9\sqrt{2}}{3\left(\sqrt{27}-3\right)}-\frac{\sqrt{150}\left(\sqrt{27}-3\right)}{3\cdot\left(\sqrt{27}-3\right)}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{9\sqrt{2}-45\sqrt{2}+3\sqrt{150}}{9\left(\sqrt{3}-1\right)}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-36\sqrt{2}+3\sqrt{150}}{9\sqrt{6}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{\sqrt{54}\cdot\left(5-4\sqrt{3}\right)}{\sqrt{486}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{5-4\sqrt{3}}{3\sqrt{3}-3}\)

\(=\frac{-7+\sqrt{3}}{6}\)(đpcm)

Answer 2

1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))

\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)

2/Ta có :

\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\sqrt{3}-3}-\frac{5\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\left(\sqrt{3}-1\right)}-\frac{5\sqrt{6}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}\right).\frac{1}{\sqrt{6}}\)

\(=\frac{3\sqrt{2}-15\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-7+\sqrt{3}}{6}\)

Vậy...