Bài 1:
\(1,=10\left(a^2-2ab+b^2\right)=10\left(a-b\right)^2\\ 2,=\left(x-5\right)^2-y^2=\left(x-y-5\right)\left(x+y-5\right)\\ 3,=7\left(x-y\right)-\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(7-x-y\right)\\ 4,=4\left[\left(x+y\right)^2-9\right]=4\left(x+y-3\right)\left(x+y+3\right)\\ 5,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ 6,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
Bài 2:
\(1,\Leftrightarrow8x-12-5x+5=5\Leftrightarrow3x=12\Leftrightarrow x=4\\ 2,\Leftrightarrow\left(x-2\right)\left(5x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{6}{5}\end{matrix}\right.\)
Bài 3:
\(a,=-4x^3y\\ b,=-9x+4x^2y\\ c,=\left(x^3+x^2-2x^2-2x+3x+3\right):\left(x+1\right)\\ =\left(x+1\right)\left(x^2-2x+3\right):\left(x+1\right)=x^2-2x+3\)
