Question

rút gọn các phân thức cùng mẫu với điều kiện các phân thức đều xác định

d)x^2-16/4x-x^2                                            h)x^2-xy/3xy-3y^2

i)(x+y)^2 -x^2/x+y+z                                     j)x^2+4x+3/2x+6

Answer 1

\(d,\dfrac{x^2-16}{4x-x^2}\left(x\ne4;x\ne0\right)\\ =\dfrac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\\ =\dfrac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}\\ =\dfrac{x+4}{-x}\\ h,\dfrac{x^2-xy}{3xy-3y^2}\left(x\ne y;y\ne0\right)\\ =\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}\\ =\dfrac{x}{3y}\\ i,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\\ =\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}\\ =x+y-z\\ j,\dfrac{x^2+4x+3}{2x+6}\left(x\ne-3\right)\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{2\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}\\ =\dfrac{x+1}{2}\)