= 0 bạn nhé

\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

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

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

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

\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{2^2-\left(\sqrt{3}\right)^2}=\sqrt{4-3}=1\)

Vì \(9>5\)\(\Rightarrow\sqrt{9}>\sqrt{5}\)\(\Rightarrow3>\sqrt{5}\)\(\Rightarrow3-\sqrt{5}>0\)

mà \(3+\sqrt{5}>0\)

\(\Rightarrow\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)

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

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

\(=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right).\left(3+\sqrt{5}\right)}\)

\(=\sqrt{4.\left(3-\sqrt{5}\right)}+\sqrt{4.\left(3+\sqrt{5}\right)}\)

\(=2.\sqrt{3-\sqrt{5}}+2.\sqrt{3+\sqrt{5}}\)

\(\sqrt{n^2+n^2\left(n+1\right)^2+\left(n+1\right)^2}\)

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

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

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

Suy ra đpcm

a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)

\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)

\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)

\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )

\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)

^{GHJHGJYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY}

\(x_1=\frac{-\left(m-1\right)+\sqrt{\left(m-1\right)^2+8}}{2a}\)

\(x_2=\frac{-\left(m-1\right)-\sqrt{\left(m-1\right)^2+8}}{2a}\)

Ta có: \(\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)

     \(=\left(\sqrt{2}.\sqrt{4+\sqrt{15}}\right).\left(\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

     \(=\sqrt{8+2\sqrt{15}}.\left(16-15\right).\left(\sqrt{5}-\sqrt{3}\right)\)

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

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

     \(=\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

     \(=5-3=2\)

Học tốt nha ^_^

Từ \(\sin\alpha.\cos\alpha=\frac{\sqrt{3}}{4}\)

\(\Rightarrow\left(\sin\alpha.\cos\alpha\right)^2=\left(\frac{\sqrt{3}}{4}\right)^2\)

\(\Rightarrow\sin^2\alpha.\cos^2\alpha=\frac{3}{16}\)

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

\(\Leftrightarrow\sin^4\alpha+2\sin^2\alpha.\cos^2\alpha+\cos^4\alpha=1\)

\(\Leftrightarrow\sin^4\alpha+\cos^4\alpha+2.\frac{3}{16}=1\)

\(\Leftrightarrow\sin^4\alpha+\cos^4\alpha+\frac{3}{8}=1\)

\(\Leftrightarrow\sin^4\alpha+\cos^4\alpha=\frac{5}{8}\)

hay \(P=\sin^4\alpha+\cos^4\alpha=\frac{5}{8}\)