Bài 1. 

a) \(cot70^o=tan20^o,cot55^o=tan35^o,cot40^o=tan50^o\)

Sắp xếp theo thứ tự tăng dần: \(tan20^o,tan28^o,tan33^o,tan35^o,tan50^o\)

b) \(sin^220^o+sin^270^o+2tan35^otan55^o\)

\(=sin^220^o+cos^220^o+2tan35^ocot35^o\)

\(=1+2=3\)

Bài 2. 

a) Xét tam giác \(ABC\)vuông tại \(A\): 

\(AH^2=BH.CH\Leftrightarrow AH=\sqrt{2.4}=2\sqrt{2}\left(cm\right)\)

\(BC=BH+CH=4+2=6\left(cm\right)\)

\(AB^2=BH.BC\Leftrightarrow AB=\sqrt{BH.BC}=\sqrt{4.6}=2\sqrt{6}\left(cm\right)\)

\(AC^2=CH.CB\Leftrightarrow AC=\sqrt{2.6}=2\sqrt{3}\left(cm\right)\)

\(\widehat{ABC}=arcsin\frac{2\sqrt{3}}{6}\approx35^o\)

b) Xét tam giác \(AHC\)vuông tại \(H\)đường cao \(HE\):

\(CH^2=AC.CE\)

Xét tam giác \(ABC\)vuông tại \(A\)đường cao \(AH\):

\(AC^2=CH.BC\)

\(\Leftrightarrow AC^4=CH^2.BC^2=AC.CE.BC^2\)

\(\Leftrightarrow AC^3=CE.BC\)

c) \(BC=\frac{AC^2}{HC}\)

\(\Leftrightarrow\frac{BC}{HC}=\frac{BC^2}{AC^2}\)

\(\Leftrightarrow\frac{2MC}{HC}=\frac{4MA^2}{AC^2}\)

\(\Leftrightarrow\frac{MH}{HC}+1=2\left(\frac{MA}{AC}\right)^2\)

\(\sqrt{2x+2\sqrt{x^2-4}}\)

\(\sqrt{x-2+2\sqrt{x-2}\sqrt{x+2}+x+2}\)\(\)

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

\(\left|\sqrt{x-2}\right|+\left|\sqrt{x+2}\right|\)

\(\sqrt{x-2}+\sqrt{x+2}\)

\(=\sqrt{7}\)

\(=\frac{6\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{4\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{100\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)

\(=\frac{6-6\sqrt{x}-4-4\sqrt{x}+100\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\frac{90\sqrt{x}-2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\frac{2\left(45\sqrt{x}-1\right)}{1-x}\)

\(x\ge\frac{3}{2}\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\)

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

dễ thấy \(\sqrt{2x+3}>0\)

\(\sqrt{2x-3}=2\)

\(2x-3=4\)

\(x=\frac{7}{2}\left(TM\right)\)
\(b,ĐKXĐ:x\ge\frac{3}{5}\)

\(\sqrt{25x^2-9}=2\sqrt{5x-3}\)

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

\(\sqrt{5x-3}\ge0\)

\(TH1:\sqrt{5x-3}=0\)

\(x=\frac{3}{5}\)

\(\sqrt{5.\frac{3}{5}+3}.0=2.0\)pt (luôn đúng<=> vô số nghiệm)

\(TH2:\sqrt{5x-3}>0\)

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

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

\(x=\frac{1}{5}\left(KTM\right)\)vì \(ĐKXĐ:x\ge\frac{3}{5}\)

\(ĐKXĐ:x\ne1;x\ge0\)

\(a,P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-2\sqrt{x}+x}{2}\right)\)

\(P=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{x-2\sqrt{x}+1}{2}\)

\(P=\frac{-2x-2\sqrt{x}}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)

\(P=\frac{-x-\sqrt{x}}{x+2\sqrt{x}+1}.x-1\)

\(P=\frac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(P=\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}\)

b,dễ thấy \(\sqrt{x}+1>0\left(\forall x\right)\)

\(< =>\sqrt{x}-1>0\)

\(x>1\)

\(c,P=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)

\(P=\frac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\sqrt{x}+1}\)

\(P=\frac{\sqrt{x}-1-x+\sqrt{x}}{\sqrt{x}+1}\)

\(P=\frac{-1+2\sqrt{x}-x}{\sqrt{x}+1}\)

\(P=\frac{-\left(x+2\sqrt{x}+1\right)+3\sqrt{x}}{\sqrt{x}+1}\)

\(P=\frac{-\left(\sqrt{x}+1\right)^2+3\sqrt{x}}{\sqrt{x}+1}\)

\(P=-\left(\sqrt{x}+1\right)+\frac{3\sqrt{x}}{\sqrt{x}+1}\)

\(P=-\left(\sqrt{x}+1\right)+\frac{3\sqrt{x}+3}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)

\(P=3-\left(\sqrt{x}+1+\frac{3}{\sqrt{x}+1}\right)\)

\(\sqrt{x}+1+\frac{3}{\sqrt{x}+1}\ge\sqrt{\sqrt{x}+1.\frac{3}{\sqrt{x}+1}}=\sqrt{3}\)

\(P\le3-\sqrt{3}\)dấu "=" xảy ra khi và chỉ khi \(\sqrt{x}+1=\frac{3}{\sqrt{x}+1}\)

\(\sqrt{x}+1=\sqrt{3}\)

\(\sqrt{x}=\sqrt{3}-1\)

\(x=3+1-2\sqrt{3}=4-2\sqrt{3}\)

\(< =>MAX:P=\sqrt{3}\)

ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

a) \(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\cdot\frac{\left(x-1\right)^2}{2}\)

\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(=-2\left(\sqrt{x}-1\right)\)

b) Để P > 0 thì \(-2\left(\sqrt{x}-1\right)>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)

Kết hợp với ĐK => Với 0 ≤ x < 1 thì P > 0

c) Ta có : \(P=-2\left(\sqrt{x}-1\right)=-2\sqrt{x}+2\le2\forall x\ge0\)

Dấu "=" xảy ra <=> x = 0 (tm)

Vậy MaxP = 2

\(a,x>0;x\ne4,9\)

\(b,Q=\left(\frac{1}{\sqrt{x}-3}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)

\(Q=\left(\frac{\sqrt{x}-\sqrt{x}+3}{x-3\sqrt{x}}\right):\left(\frac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(Q=\frac{3}{x-3\sqrt{x}}:\frac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\frac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\)

\(Q=\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)

\(c,Q< 0< =>\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)

\(-5\sqrt{x}< 0\)

\(< =>3\sqrt{x}-6>0\)

\(\sqrt{x}>2\)

\(x>4\)

\(1+\sqrt{2x}-x^2+1\)

\(2+\sqrt{2x}-x^2\)

\(-\left[x^2+\sqrt{2x}+\left(\frac{\sqrt{2}}{2}\right)^2\right]+\frac{5}{2}\)

\(-\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{5}{2}\le\frac{5}{2}\)dấu "=" xảy ra khi và chỉ khi \(x=-\frac{\sqrt{2}}{2}\)

\(< ==>MAX=\frac{5}{2}\)