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a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
\(X^2-2XY+Y^2+2X-2Y\)
\(\Leftrightarrow\left(X^2-2XY+Y^2\right)+\left(2X-2Y\right)\)
\(\Leftrightarrow\left(X-Y\right)^2+2\left(X-Y\right)\)
\(\Leftrightarrow\left(X-Y\right)\left(X-Y+2\right)\)
Tk mình nhé.
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y\right)+y^2\left(y-x\right)+z^2\left(x-y\right)\)
\(=\left[x^2\left(y-z\right)-y^2\left(y-z\right)\right]+\left[y^2\left(y-x\right)-z^2\left(y-x\right)\right]\)
\(=\left(x^2-y^2\right)\left(y-z\right)+\left(y^2-z^2\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)+\left(y-z\right)\left(y+z\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x+y\right)-\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
a) \(-10x^3+2x^2=0\)
\(\Rightarrow-2x^2\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(5x\left(x-2016\right)-x+2016=0\)
\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)
\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)
a: Ta có: \(-10x^3+2x^2=0\)
\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
2x2+x-3
=2x2+3x-2x-3
=2x(2x+3)-(2x+3)
=(2x+3)(2x-1)
Chúc bạn học tốt
2x2+x-3
=2𝑥2+3𝑥−2𝑥−3
=𝑥(2𝑥+3)−1(2𝑥+3)
=(𝑥−1)(2𝑥+3)
ez mà bro :D