Phân thức đại số
Các câu hỏi tương tự
Kết quả của phép tính \(\dfrac{x^2-4}{x^2-2x}\) . \(\dfrac{-3}{x+2}\) là :
Thực hiện phép tính:
a) \(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
b) \(\dfrac{x}{x-3}+\dfrac{-9}{x^2-3x}\)
c) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{x-2}{x+2}\)
e) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^2}\)
Kết quả phép tính \(\dfrac{3x-20}{x-5}\) + \(\dfrac{2x-5}{x-5}\) là :
A : 5
B : 5x
C : 3
D : 3 (x + 2)
Kết quả của phép chia \(\dfrac{x^2+x}{5x^2-10x+5}\) : \(\dfrac{3x+3}{5x-5}\) là :
\(\dfrac{x-14}{x^2-4x}-\dfrac{3}{2x}+\dfrac{x+1}{2x-8}\)
Thực hiện phép tính :
a/ (x - 1)^2 - (4x + 3) (2 - x)
b/ (15x^3y^2 - 6x^2y^3) : 3x^2y^2 = (15x^3y^2 : 3x^2y^2) - (6x^2y^3 : 3x^2y^2) = 5x - 2y
c/\(\dfrac{x+7}{x-7}\) - \(\dfrac{x-7}{x+7}\) +\(\dfrac{4x^2}{x^2-49}\)
Kết quả thu gọn phân thức \(\dfrac{x^2-4x+4}{x^2-4}\) là :
a) dfrac{4x}{x^2+2x}+dfrac{8}{x^2+2x} b) dfrac{2x-3x}{x-2}-dfrac{2x-4}{x-2} c) dfrac{2x-1}{x+3}-dfrac{3x+2}{x+3}d) dfrac{11x}{2x-3}-dfrac{18-x}{2x-3}e) dfrac{3left(x-2right)}{2x+1}-dfrac{9x-3}{2x+1}
Đọc tiếp
a) \(\dfrac{4x}{x^2+2x}\)+\(\dfrac{8}{x^2+2x}\)
b) \(\dfrac{2x-3x}{x-2}\)-\(\dfrac{2x-4}{x-2}\)
c) \(\dfrac{2x-1}{x+3}\)-\(\dfrac{3x+2}{x+3}\)
d) \(\dfrac{11x}{2x-3}\)-\(\dfrac{18-x}{2x-3}\)
e) \(\dfrac{3\left(x-2\right)}{2x+1}\)-\(\dfrac{9x-3}{2x+1}\)
chứng minh rằng :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b)left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3x}-x-1right)right]:dfrac{x+1}{x}dfrac{2x}{x-1}
c)left[dfrac{2}{left(x+1right)^3}left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
Đọc tiếp
chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)