Bài 1:
\(a,=5\left(x-2y\right)+3x\left(x-2y\right)=\left(3x+5\right)\left(x-2y\right)\\ b,=\left(x-5\right)^2-y^2=\left(x-y-5\right)\left(x+y-5\right)\\ c,=x^2+6x-x-6=\left(x+6\right)\left(x-1\right)\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-5\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\\ c,\Leftrightarrow\left(x^3+x^2-x^2-x+2x+2+3\right)⋮\left(x+1\right)\\ \Leftrightarrow\left[\left(x+1\right)\left(x^2-x+2\right)+3\right]⋮\left(x+1\right)\\ \Leftrightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)
Bài 1:
a,=5(x−2y)+3x(x−2y)=(3x+5)(x−2y)b,=(x−5)2−y2=(x−y−5)(x+y−5)c,=x2+6x−x−6=(x+6)(x−1)a,=5(x−2y)+3x(x−2y)=(3x+5)(x−2y)b,=(x−5)2−y2=(x−y−5)(x+y−5)c,=x2+6x−x−6=(x+6)(x−1)
Bài 2:
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