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28.
\(\Leftrightarrow2x+1=x-1\)
\(\Leftrightarrow x=-2\)
29.
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x\left(x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)
30.\(\Leftrightarrow3x^2+3x+2x+2=0\)
\(\Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(28,\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=-x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ 29,\\ \Leftrightarrow x\left(x^2-5x+6\right)=0\\ \Leftrightarrow x\left(x^2-2x-3x+6\right)=0\\ \Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(30,\\ \Leftrightarrow3x^2+3x+2x+2=0\\ \Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\end{matrix}\right.\)
\(\Rightarrow12+20x-60=45x-15\Leftrightarrow25x=-33\Leftrightarrow x=-\dfrac{33}{25}\)
a) \(A=9x^2+5x+1\)
\(A=9x^2+5x+\frac{25}{36}+\frac{11}{36}\)
\(A=\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\)
Có: \(\left(3x+\frac{5}{6}\right)^2\ge0\)
\(\Rightarrow\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}\)
Dấu = xảy ra khi: \(\left(3x+\frac{5}{6}\right)^2=0\Rightarrow3x+\frac{5}{6}=0\)
\(\Rightarrow x=-\frac{5}{18}\)
Vậy: \(Min_A=\frac{11}{36}\) tại \(x=-\frac{5}{18}\)
b) \(B=4x^2+12x-8\)
\(B=4x^2+12x+9-17\)
\(B=\left(2x+3\right)^2-17\)
Có: \(\left(2x+3\right)^2\ge0\)
\(\Rightarrow\left(2x+3\right)^2-17\ge-17\)
Dấu = xảy ra khi: \(\left(2x+3\right)^2=0\Rightarrow2x+3=0\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy: \(Min_B=-17\) tại \(x=-\frac{3}{2}\)
a. 2x2.(3x3 + 2x)
= 2x2.3x3 + 2x2.2x
= 6x5 + 4x3
b. 3x.(x2 + 2x + 2)
= 3x.x2 + 3x.2x + 3x.2
= 3x3 + 6x2 + 6x
