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

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

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

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

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

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

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

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

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

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

@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

\(ĐKXĐ:x\ne2;4\)

\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=3\frac{1}{5}\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=\frac{16}{5}\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow x^2-7x+12+x^2-4x+4=\frac{16}{5}\left(x^2-6x+8\right)\)

\(\Leftrightarrow2x^2-11x+16=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)

\(\Leftrightarrow\frac{6}{5}x^2-\frac{41}{5}x+\frac{48}{5}=0\)

\(\Leftrightarrow6x^2-41x+48=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{16}{3}\\x=\frac{3}{2}\end{cases}}\)

rõ hơn đi bạn