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 )

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

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

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

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

1. Ta có: \(9< 10\)\(\Rightarrow\sqrt{9}< \sqrt{10}\)\(\Rightarrow3< \sqrt{10}\)\(\Rightarrow3-\sqrt{10}< 0\)(1)

Vì \(3< \sqrt{10}\)\(\Rightarrow2.3< 2\sqrt{10}\)\(\Rightarrow6< 2\sqrt{10}\)\(\Rightarrow2\sqrt{10}-6>0\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{\left(3-\sqrt{10}\right)^2}+\sqrt{\left(2\sqrt{10}-6\right)^2}\)

\(=\left|3-\sqrt{10}\right|+\left|2\sqrt{10}-6\right|\)

\(=\sqrt{10}-3+2\sqrt{10}-6=3\sqrt{10}-9\)

4. Vì \(x>0\)\(\Rightarrow x.\sqrt{\frac{9}{x}}+5\sqrt{x}=\sqrt{x^2.\frac{9}{x}}+5\sqrt{x}=\sqrt{9x}+5\sqrt{x}\)

\(=3\sqrt{x}+5\sqrt{x}=8\sqrt{x}\)

Thiếu 1 phương trình :

\(4x^2-4\left(2n+1\right)x+4n^2+96mnp+1=0\)

Định lí PYTAGO cho tam giác ABC vuông tại A: \(BC^2=AB^2+AC^2=2AB^2\Rightarrow BC=AB\sqrt{2}\)

Xét tam giác ABC vuông tại A: \(sinB=\frac{AC}{BC}=\frac{AB}{AB\sqrt{2}}=\frac{\sqrt{2}}{2}\)\(cosB=\frac{AB}{BC}=\frac{AB}{AB\sqrt{2}}=\frac{\sqrt{2}}{2}\)

\(tanB=\frac{AC}{AB}=\frac{AB}{AB}=1\), \(cotB=\frac{AB}{AC}=\frac{AB}{AB}=1\)

Vì tam giác ABC vuông cân tại A-->B=45⁰

Vậy \(sin45^0=cos45^0\frac{\sqrt{2}}{2},tan45^0=cot45^0=1\)

,

Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\)   \(\left(ĐK:x\ge1\right)\)

    \(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)

    \(\Leftrightarrow4.\left(x-1\right)=8x+12\)

    \(\Leftrightarrow4x-4=8x+12\)

    \(\Leftrightarrow-4x=16\)

    \(\Leftrightarrow x=-4\left(L\right)\)

Vậy \(S=\varnothing\)

\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\) 

\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\) 

\(2\sqrt{x-1}=2\sqrt{2x+3}\) 

\(\sqrt{x-1}=\sqrt{2x+3}\) 

\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\) 

\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\) 

\(\Rightarrow x=2\)

Ta có: \(5a\sqrt{64ab^3}-\sqrt{3}.\sqrt{12a^3b^3}-5b\sqrt{81a^3b}\)

    \(=5a.8b.\sqrt{ab}-\sqrt{3}.2\sqrt{3}.ab.\sqrt{ab}-5b.9a\sqrt{ab}\)

    \(=40ab.\sqrt{ab}-6ab.\sqrt{ab}-45ab.\sqrt{ab}\)

    \(=40ab.\sqrt{ab}-6ab.\sqrt{ab}-45ab.\sqrt{ab}\)

    \(=-11ab\sqrt{ab}\)

\(5a\sqrt{64ab^3}\) \(-\sqrt{3}\)\(.\sqrt{12a^3b^3}\)\(-5b\sqrt{81a^3b}\)

\(=40ab\sqrt{ab}\)\(-6ab\sqrt{ab}-45ab\sqrt{ab}\)

\(=\left(40ab-6ab-45ab\right)\sqrt{ab}\)

\(=-11ab\sqrt{ab}\)