\(^{x^2-3x+2}\)

\(=x^2-x-2x+2\)

\(=\left(x^2-x\right)-\left(2x-2\right)\)

\(=x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

\(x^2-3x+2=0\)

\(x^2-4x+2^2+x-2=0\)

\(\left(x-2\right)^2+\left(x-2\right)=0\)

\(\left(x-2\right)\left(x-2+1\right)=0\)

\(\left(x-2\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}}\)

Vậy ..........

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

<=>\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)

<=>\(ab+bc+ca=0\)

<=>\(\frac{ab+bc+ca}{abc}=0\)

<=> \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)

<=>\(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

<=>\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c}^3\)

<=>\(\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=\frac{-1}{c}^3\)

<=>\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Ta có: \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{3abc}{abc}=3\)

Câu hỏi của Nàng tiên cá - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

giúc mk với 

1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)

ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)

<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)

<=> \(\frac{3x+10}{x^2+2x-3}=0\)

<=> \(3x+10=0\)

<=> \(x=-\frac{10}{3}\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

                                    Vậy S = { 0, 6}