áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

P \(\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}=\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\)

mà \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=2\)=> P\(\ge\frac{\left(2+2\right)^2}{2}=8\)

P đạt GTNN khi a=b =1

\(5x^2+7x-6=0\)

\(5x^2-10x+3x-6=0\)

\(5x\left(x-2\right)+3\left(x-2\right)=0\)

\(\left(x-2\right)\left(5x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\5x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-\frac{3}{5}\end{cases}}}\)

\(5x^2+7x-6=0\)

Ta có: \(\Delta=7^2+4.5.6=169,\sqrt{\Delta}=13\)

Vậy pt có 2 nghiệm phân biệt:

\(x_1=\frac{-7+13}{10}=\frac{3}{5}\)'\(x_1=\frac{-7-13}{10}=-2\)

\(5x^2+7x-6=0\)

\(\Leftrightarrow5x^2-3x+10x-6=0\)

\(\Leftrightarrow x\left(5x-3\right)+2\left(5x-3\right)=0\)

\(\Leftrightarrow\left(5x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-2\end{cases}}\)

Vậy ...

\(5x^2+7x-6=0\)

\(\Leftrightarrow5x^2-3x+10x-6=0\)

\(\Leftrightarrow\left(5x^2+3x\right)+\left(10x-6\right)=0\)

\(\Leftrightarrow x\left(5x-3\right)+2\left(5x-3\right)=0\)

\(\Leftrightarrow\left(5x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-2\end{cases}}}\)

với m\(\ge n;p\ge q\)=> (m-n)(p-q) \(\ge0\)<=> mp+nq \(\ge mq+np\)<=> mp+ nq\(\ge\frac{1}{2}\left(m+n\right)\left(p+q\right)\)

giả sử \(a\ge b=>\frac{1}{b+1}\ge\frac{1}{a+1};\)áp dụng bdt trên ta được

\(\frac{a}{b+1}+\frac{b}{a+1}\ge\frac{1}{2}\left(a+b\right)\left(\frac{1}{b+1}+\frac{1}{a+1}\right)\ge\frac{1}{2}\left(a+b\right)\frac{4}{a+1+b+1}\)( theo bdt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\))

vậy \(\frac{a}{b+1}+\frac{b}{a+1}+\frac{1}{a+b}\ge\frac{2\left(a+b\right)}{a+b+2}+\frac{1}{a+b}\)đặt a+b=X

ta được \(\frac{2X}{X+2}+\frac{1}{X}=\frac{2X^2+X+2}{\left(X+2\right)X}\ge\frac{3}{2}< =>4X^2+2X+4\ge3X\left(X+2\right)< =>\)(X-2)² \(\ge0\)(đúng)

dấu '=' sảy ra khi X = a+b=2 và a=b hay a = b =1

n(n^3+2n^2-n-2)

n(n-2)(n-1)(n+1)

\(n^4+2n^3-n^2-2n\)

\(=n\left(n^3+2n^2-n-2\right)\)

\(=n\left[n^2\left(n+2\right)-\left(n+2\right)\right]\)

\(=n\left[\left(n^2-1\right)\left(n+2\right)\right]\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

Tích 4 số liên tiếp chia hết cho 4 nên \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮4\)

Dễ c/m \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮2\)

và \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮3\)

Mà (2,3,4) = 1 nên \(n^4+2n^3-n^2-2n⋮24\left(đpcm\right)\)