c) Áp dụng BĐT cô si cho 2 hai số dương \(a;b\) ta có:

\(a+b\ge2\sqrt{ab}\)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{1}{\sqrt{ab}}\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Dấu "=" xảy ra khi \(\Leftrightarrow a=b\)

\(\left(x-3\right)\left(x+5\right)+20=x^2+2x-15+20=x^2+2x+5=\left(x+1\right)^2+4\)

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+4\ge4\Rightarrow\left(x-3\right)\left(x+5\right)+20\ge4\)

Dấu "=" xảy ra khi (x+1)²=0 =>x+1=0=>x=-1

Nguyên trang bất đăng thức Bunhacoxki  rồi. 

ảnh đại diện của mình đẹp ko

a) ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

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

\(=\frac{x^3-3}{\left(x+1\right)\left(x-3\right)}-\frac{2x-6}{x+1}-\frac{x+3}{x-3}\)

\(=\frac{x^3-3}{\left(x+1\right)\left(x-3\right)}-\frac{\left(2x-6\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^3-3-\left(2x^2-12x+18\right)-\left(x^2+4x+3\right)}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^3-3x^2+8x-24}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^2\left(x-3\right)+8\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}=\frac{\left(x-3\right)\left(x^2+8\right)}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^2+8}{x+1}\)

b) Để P nguyên thì \(x^2+8⋮x+1\)

\(\Leftrightarrow x^2+2x+1-2x+7⋮x+1\)

\(\Leftrightarrow\left(x+1\right)^2-2x+7⋮x+1\)

mà \(\left(x+1\right)^2⋮x+1\)

nên \(-2x+7⋮x+1\)

\(\Leftrightarrow-2x-2+9⋮x+1\)

mà \(-2x-2⋮x+1\)

nên \(9⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(9\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Leftrightarrow x\in\left\{0;-2;2;-4;8;-10\right\}\)(tm)

Vậy: Khi \(x\in\left\{0;-2;2;-4;8;-10\right\}\) thì P có giá trị nguyên

c)

Khi x>-1 thì x+1>0

mà \(x^2+8\ge0\forall x\)

nên khi x>-1 và \(x\ne3\) thì \(P=\frac{x^2+8}{x+1}>0\)

Để \(P\ge4\) thì \(\frac{x^2+8}{x+1}\ge4\)

\(\Leftrightarrow x^2+8\ge\left(x+1\right)\cdot4\)

\(\Leftrightarrow x^2+8\ge4x+4\)

\(\Leftrightarrow x^2+8-4x-4\ge0\)

\(\Leftrightarrow x^2-4x+4\ge0\)

\(\Leftrightarrow\left(x-2\right)^2\ge0\)(luôn đúng)

Dấu '=' xảy ra khi x-2=0

hay x=2