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b) Ta có: \(x^3-x^2y-xy^2+y^3\)
\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)^2\)
\(3x-3y+a\left(x-y\right)=\left(x-y\right)\left(a+3\right)\)
Câu 1:
\(4x^2+16x-9\)
\(=4x^2+18x-2x-9\)
\(=2x\left(2x+9\right)-\left(2x+9\right)\)
\(=\left(2x-1\right)\left(2x+9\right)\)
Câu 2:
\(6x^2-11x+3=0\)
\(\Leftrightarrow6x^2-2x-9x+3=0\)
\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(=3x\cdot\left(x-5y\right)+2y\cdot\left(x-5y\right)\)
\(=\left(x-5y\right)\left(3x+2y\right)\)
\(3x\left(x-5y\right)-2y\left(5y-x\right)\)
\(=3x\left(x-5y\right)+2y\left(x-5y\right)\)
\(=\left(x-5y\right)\left(3x+2y\right)\)
\(-3x^2+4x-2020\)
\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)
\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)
\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)
\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)
(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))
3x^3+x^2 -15x^2-5x+9x+3
= (3x+1)(x^2-5x+3)
\(3x^3-14x^2+4x+3\)
\(=3x^3+x^2-15x^2-5x+9x+3\)
\(=x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-5x+3\right)\)