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a) Ta có: |2x-3|=x-6
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-6\\2x-3=6-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3-x+6=0\\2x-3-6+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;3\right\}\)
Ta có:
\(M=\frac{19a+3}{1+b^2}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)
\(=19a-\frac{19ab^2-3}{b^2+1}+19b-\frac{19bc^2-3}{c^2+1}+\frac{19ca^2-3}{a^2+1}\)
\(\ge19\left(a+b+c\right)-\frac{19ab^2-3}{2b}-\frac{19bc^2-3}{2c}-\frac{19ca^2-3}{2a}\)
\(=19\left(a+b+c\right)-19\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\ge19.3-\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Lại có:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\frac{\left(1+1+1\right)^2}{ab+bc+ca}=\frac{3.9}{3}=9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
\(\Rightarrow M\ge\frac{19.3}{2}+\frac{3}{2}.3=33\)
\(\)
bài 3:
b. x^3-6x^2+12x-8+6(x^2+2x+1)-(x^3+3x^2+9x-3x^2-9x-27)=97
=>x^3-6x^2+12x-8+6x^2+12x+6-x^3+27=97
=>24x+25=97
=>x=3