2) ĐK : x ∈ R

\(\Leftrightarrow\sqrt{4x^2-6x+5}+4x^2-6x-7=0\)(1)

Đặt \(\sqrt{4x^2-6x+5}=a\left(a\ge0\right)\)

(1) trở thành a² + a - 12 = 0 <=> ( a - 3 )( a + 4 ) = 0 <=> a = 3 ( tm ) hoặc a = -4 ( ktm )

\(\Rightarrow\sqrt{4x^2-6x+5}=3\Leftrightarrow2x^2-3x-2=0\Leftrightarrow\left(x-2\right)\left(2x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}\left(tm\right)}\)

Vậy phương trình đã cho có tập nghiệm S = { 2 ; -1/2 }

\(\sqrt{2}x-2+2\sqrt{2}x-3+\sqrt{2}x+13+8\sqrt{2}x-3=5\)

\(\Leftrightarrow\left(\sqrt{2}x+2\sqrt{2}x+\sqrt{2}x+8\sqrt{2}x\right)-2-3+13-3=5\)

\(\Leftrightarrow12\sqrt{2}x+5=5\)

\(\Leftrightarrow12\sqrt{2}x=0\)

\(\Leftrightarrow x=0\)

Vậy tập nghiệm của phương trình là S={0}

\(ĐKXĐ:x\ge0;x\ne\frac{1}{9}\)

\(\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(\left(\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(\frac{3x-3\sqrt{x}+\sqrt{x}-1+5\sqrt{x}+1}{9x-1}.\frac{3\sqrt{x}+1}{3}\)

\(\frac{3x+3\sqrt{x}}{9x-1}.\frac{3\sqrt{x}+1}{3}\)

\(\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

\(5+2\sqrt{7}\)và \(3\sqrt{10}\)đề bài thế này hả bạn

\(\left(5+2\sqrt{7}\right)^2=25+28+20\sqrt{7}\)

\(=53+20\sqrt{7}\)

\(\left(3\sqrt{10}\right)^2=90=53+37\)

so sánh \(20\sqrt{7}\)và \(37\)

\(\left(20\sqrt{7}\right)^2=2800\)

\(37^2=1369< 2800\)

\(< =>\left(5+2\sqrt{7}\right)^2>3\sqrt{10}^2\)

\(5+2\sqrt{7}>3\sqrt{10}\)

\(\left(5-2\sqrt{7}\right)^2=53-20\sqrt{7}\)

\(\left(3-\sqrt{10}\right)^2=19-6\sqrt{10}=53-\left(34+6\sqrt{10}\right)\)

so sánh \(20\sqrt{7}\)và \(34+6\sqrt{10}\)

\(\left(20\sqrt{7}\right)^2=2800=1516+1284\)

\(\left(34+6\sqrt{10}\right)^2=1516+408\sqrt{10}\)

\(\left(1284\right)^2< \left(408\sqrt{10}\right)^2\)

\(1284< 408\sqrt{10}< =>20\sqrt{7}< 34+6\sqrt{10}\)

\(< =>\left(5-2\sqrt{7}\right)^2>\left(3-\sqrt{10}\right)^2\)

dễ thấy \(5-2\sqrt{7}< 0\)

\(3-\sqrt{10}< 0\)

vậy bình nên cái nào lớn hơn thì cái đó bé hơn

\(< =>5-2\sqrt{7}< 3-\sqrt{10}\)

cos20,sin65,cos28,sin40,cos88 

Giải thích các bước giải:

 đổi sin40=cos(90-40)=cos50

sin65=cos(90-65)=cos25

\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)

\(\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)

\(\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|}{\sqrt{2}}\)

\(\frac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}\)

\(=\sqrt{2}\)

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.