rút gọn phân thức a)(x^2-6x+9)/(4x^2-12x) b)(10xy^2-5x^2)/(2x^2-8y^2) c)(x^2+7x+12)/(x^2+5x+4) d)(5x^2+10xy+5y^2)/(3y^2-x^2)

Đáp án:

 `a)`

$\dfrac{x^2-6x+9}{4x^2-12x}$ $\text{(x $\neq$ 0;3)}$

$= \dfrac{x^2-2.x.3+3^2}{4x.(x-3)}$

$=\dfrac{(x-3)^2}{4x.(x-3)}$

$= \dfrac{x-3}{4x}$

`b)`

$\dfrac{10xy^2-5x^2}{2x^2-8y^2}$

$=\dfrac{5x.(2y^2-x)}{2.(x^2-4y^2)}$

$=\dfrac{5x.(2y^2-(\sqrt{x})^{2}}{2.(x-2y).(x+2y)}$

$=\dfrac{5x.(2y-\sqrt{x}).(2y+\sqrt{x})}{2.(x-2y).(x+2y)}$ (không thể rút gọn được nữa)

sửa đề cho hợp lí (nếu có thể):

$\dfrac{10y^2-5x^2}{2x^2-8y^2}$

$=\dfrac{5.(2y^2-x^2)}{2.(x^2-4y^2)}$

$=\dfrac{-5.(x-2y).(x+2y)}{2.(x-2y).(x+2y)}$

$=-\dfrac{5}{2}$

`c)`

 $\dfrac{x^2+7x+12}{x^2+5x+4}$ $\text{(x $\neq$ -4;-1)}$

$=\dfrac{x^2+3x+4x+12}{x^2+x+4x+4}$

$=\dfrac{x.(x+3)+4.(x+3)}{x.(x+1)+4.(x+1)}$

$=\dfrac{(x+4).(x+3)}{(x+4).(x+1)}$

$=\dfrac{x+3}{x+1}$

`d)`

$\dfrac{5x^2+10xy+5y^2}{3y^2-x^{2}}$

$=\dfrac{5.(x^2+2xy+y^2)}{(\sqrt{3y^2})^{2}-x^{2}}$

$=\dfrac{5.(x+y)^{2}}{(\sqrt{3y^2}-x).(\sqrt{3y^2}+x)}$

sửa đềcho hợp lí:

$\dfrac{5x^2+10xy+5y^2}{3y^2-3x^2}$

$=\dfrac{5.(x^2+2xy+y^2)}{3.(y^2-x^2)}$

$=\dfrac{5.(x+y)^{2}}{-3(x-y).(x+y)}$

$=-\dfrac{5.(x+y)}{3.(x-y)}$