Rút gọn các phân thức: 20 x 2 - 45 2 x + 3 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x khac +-3
A=\(\hept{\begin{cases}\\\end{cases}\frac{xI\left(x-3\right)I}{5x^2-45}=\frac{xI\left(x-3\right)I}{5\left(x^2-3^2\right)}}\)
\(\frac{xIx-3I\overline{ }}{5\left(x-3\right)\left(x+3\right)^{ }_{ }}\)
x>3 A=\(\frac{x}{5\left(x+3\right)}\)
x<3 A=-\(\frac{x}{5\left(x+3\right)}\)
\(1,\Leftrightarrow x=10\\ 2,=x^2-4x+4-x^2+4x=4\\ 3,=\left(x+y\right)^2-49=\left(x+y+7\right)\left(x+y-7\right)\)
\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)
b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)
a)\(\dfrac{x^2-4xy+4y^2}{xy-2y^2}\)
=\(\dfrac{x^2-4xy+\left(2y\right)^2}{y\left(x-2y\right)}\)
=\(\dfrac{\left(x-2y\right)^2}{y\left(x-2y\right)}\)
=\(\dfrac{x-2y}{y}\)
b)\(\dfrac{x^3-36x}{x^2+6x}\)
=\(\dfrac{x\left(x^2-6^2\right)}{x\left(x+6\right)}\)
=\(\dfrac{x\left(x+6\right)\left(x-6\right)}{x\left(x+6\right)}\)
= \(x-6\)
#Fiona
Chúc bạn học tốt !
a) \(\dfrac{3x^2+6xy}{6x^2}=\dfrac{3x\left(x+2y\right)}{6x^2}=\dfrac{x+2y}{2x}\)
b) \(\dfrac{2x^2-x^3}{x^2-4}=\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x^2}{x+2}\)
c) \(=\dfrac{x+1}{x^3+1}=\dfrac{x+1}{\left(x+1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`
`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`
`= -x^2/(x+2)`
`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`
a: Thay x=-3 vào B, ta được:
\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)
b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)
ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)
Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)