Đặt \(2n+1=a^2,3n+1=b^2\).

\(15n+8=9\left(2n+1\right)-\left(3n+1\right)=9a^2-b^2=\left(3a-b\right)\left(3a+b\right)\)

Hiển nhiên \(3a+b>1\).

Nếu \(3a-b=1\Rightarrow b+1⋮3\).

mà \(b^2\equiv1\left(mod3\right)\Leftrightarrow b\equiv1\left(mod3\right)\Leftrightarrow b\equiv2\left(mod3\right)\)mâu thuẫn

do đó \(3a-b\ne1\).

Do đó \(15n+8\)là hợp số. 

\(\sqrt{96}.\sqrt{125}\)

\(\sqrt{16.6}\sqrt{25.5}\)

\(4.5\sqrt{6.5}\)

\(20\sqrt{30}\)

\(b,\sqrt{a^4b^5}\)

\(a^2b^2\sqrt{b}\)

\(c,\sqrt{a^6b^{11}}\)

\(a^3b^5\sqrt{b}\)

\(d,\sqrt{a^3\left(1-a\right)^4}\)

\(a\left(1-a\right)^2\sqrt{a}\)

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

\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}\right)\)

#H

bjfjfj

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 \(ĐKXĐ:x\ge\frac{3}{5}\)