Câu 3 : 

\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)  ĐKXđ : \(x\ne\pm1\)

\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)

\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)

\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)

\(A=\frac{10}{x+1}\)

\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)

ĐKXđ : \(x\ne0;x\ne3\)

\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)

\(x^4+3x^2-2x+3=\left(x^4-x^3+x^2\right)+\left(x^3-x^2+x\right)+\left(3x^2-3x+3\right)\)

                                        \(=x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+3\left(x^2-x+1\right)\)

                                         \(=\left(x^2+x+3\right)\left(x^2-x+1\right)\)

\(x^4+2x^3+3x^2+2x+1.\)

\(=x^4+x^3+x^3+x^2+x^2+x^2+x+x+1\)
\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)

\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x+1\right)^2+x\left(x+1\right)^2+\left(x+1\right)^2\)

\(=\left(x+1\right)^2\left(x^2+x+1\right)\)

\(=\left(x+1\right)^2\left(x+1\right)^2\)

\(=\left(x+1\right)^4\)

@wi

\(x^2+x+1=\left(x+1\right)^2???\)

\(x^2+2x+1=\left(x+1\right)^2\)chứ

Ta có  \(P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\)

\(P=\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\)

\(P=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

\(P=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

\(P=3+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

Áp dụng bdt Cô-si ( tự làm lười lắm :>)

\(\Rightarrow P=3+2+2+2=9\)

\(\Rightarrow P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\)

GTNN của P là 9

\(P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(P=\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2\right]\left[\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{c}}\right)^2\right]\)

Áp dụng BĐT Bunhiacopxki

\(\Rightarrow P\ge\left(\sqrt{a}.\frac{1}{\sqrt{a}}+\sqrt{b}.\frac{1}{\sqrt{b}}+\sqrt{c}.\frac{1}{\sqrt{c}}\right)^2=\left(1+1+1\right)^2=9\)

Vậy Min P = 9 <=> a = b = c = 1