Question

Answer 1

Bài 1:

\(1,\\ a,=\left(x+1\right)^2;b,=\left(5x-2y\right)^2\\ 2,\\ a,=5xy\left(x+2y\right);b,=x^2\left(x+3\right)-9\left(x+3\right)=\left(x-3\right)\left(x+3\right)^2\\ 3,\\ a,=\dfrac{1}{2x};b,=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)

Bài 2:

\(a,=4x^2-1+2-4x^2=1\\ b,=\left(x^3+2x^2+2x^2+4x+2x+4\right):\left(x+2\right)\\ =\left(x+2\right)\left(x^2+2x+2\right):\left(x+2\right)\\ =x^2+2x+2\\ c,=\dfrac{x^2+6+6x+3}{x+3}=\dfrac{\left(x+3\right)^2}{x+3}=x+3\\ d,=\dfrac{7-4x+3x-6+x^2-x}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(x-1\right)^2}{\left(x-2\right)\left(x-1\right)}=\dfrac{x-1}{x-2}\)