Câu 2:  \(x^2-5x+1=0\Leftrightarrow x^2-2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+1=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{21}{4}=0\Leftrightarrow x-\frac{5}{2}=\pm\frac{\sqrt{21}}{2}\)\(\Leftrightarrow x=\pm\frac{\sqrt{21}+5}{2}\)

Thay vào biểu thức đó: 

\(\frac{x^2+1}{x^2}=1+\frac{1}{x^2}=1+\frac{1}{\frac{\left(\sqrt{21}+5\right)^2}{4}}\)

\(=1+\frac{1}{\frac{21+10\sqrt{21}+25}{4}}=1+\frac{4}{46+10\sqrt{21}}=\frac{50+10\sqrt{21}}{46+10\sqrt{21}}\)

\(=\frac{25+5\sqrt{10}}{23+5\sqrt{10}}\). ĐS...

\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

\(M=x^2+xy+y^2-3x-3\)

\(=\dfrac{1}{4}x^2+xy+y^2+\dfrac{3}{4}x^2-3x-3\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x-1\right)\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x+1-2\right)\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{2}x-1\right)^2-6>=-6\forall x,y\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-1=0\\\dfrac{1}{2}x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}x=-\dfrac{1}{2}\cdot2=-1\end{matrix}\right.\)

Câu 2:

 (5x-3)=2x+1 

   <=> 5x-2x=3+1

   <=> 3x=4

   <=> x=4/3