= (sin²x )³ + (cos²x)³ + 3sin²x. cos²x = (sin²x + cos²x).(sin⁴x - sin²x.cos²x + cos⁴x) + 3sin²x. cos²x 

=  sin⁴x + 2sin²x.cos²x + cos⁴x  =  (sin²x + cos²x)² = 1² = 1

mk moi hoc lop 6 thui

chuc bn hoc gioI!n

nhaE@@

bn nhae$

Điều kiện : x > 0

pt <=> \(\left(0,5.\sqrt{2}-\sqrt{\frac{8}{25}} +\sqrt{\frac{1}{4}}\right).\sqrt{\frac{1}{x}}=\frac{1}{5}\)

<=> \(\left(\frac{\sqrt{2}}{2}-\frac{2\sqrt{2}}{5}+\frac{1}{2}\right).\sqrt{\frac{1}{x}}=\frac{1}{5}\)

<=> \(\frac{\sqrt{2}+5}{10}.\sqrt{\frac{1}{x}}=\frac{1}{5}\) <=>  \(\sqrt{\frac{1}{x}}=\frac{1}{5}:\frac{\sqrt{2}+5}{10}=\frac{2}{\sqrt{2}+5}\)

=> \(\sqrt{x}=\frac{\sqrt{2}+5}{2}\) => x =  \(\frac{\left(\sqrt{2}+5\right)^2}{4}\) (thỏa mãn)

Vậy....

\(7-2\sqrt{10}=5-2\sqrt{5}.\sqrt{2}+2=\left(\sqrt{5}-\sqrt{2}\right)^2\)

\(7+2\sqrt{10}=\left(\sqrt{5}+\sqrt{2}\right)^2\)

\(\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-\sqrt{2}}{3}\)

\(=\frac{2\sqrt{5}}{3}\)

\(\sqrt{3}+2+\sqrt{7-4\sqrt{3}}=\sqrt{3}+2+\sqrt{4-2.2\sqrt{3}+3}\)

=\(\sqrt{3}+2+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+2+2-\sqrt{3}=4\)

=>ĐPCM

\(\left(\frac{\sqrt{3}}{\sqrt{2}}+\sqrt{3}\right)-\sqrt{6}=\left(\frac{\sqrt{6}}{\sqrt{2}}+\sqrt{3}\right)-\sqrt{6}=\frac{\sqrt{6}+2\sqrt{3}}{2}-\sqrt{6}=\frac{-\sqrt{6}+2\sqrt{3}}{2}\)

Chỉ mìnhcách viết phân số đi, mình làm giùm cho. 

\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)

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

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

=\(\sqrt{m-1}+1+\sqrt{m-1}-1=2\sqrt{m-1}\)

\(\left(5\sqrt{48}+4\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)

\(=\frac{5\sqrt{48}}{\sqrt{3}}+\frac{4\sqrt{27}}{\sqrt{3}}-\frac{2\sqrt{12}}{\sqrt{3}}\)

\(=\frac{5\sqrt{3.16}+4\sqrt{3.9}-2\sqrt{3.4}}{\sqrt{3}}\)

\(=5.\sqrt{16}+4.\sqrt{9}-2\sqrt{4}\)

\(=5.4+4.3-2.2\)

\(=28\)

\(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)

=\(\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.2+4}+\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.2+4}\)

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

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