Rút gọn các phân thức: x 2 + 2 2 x + 2 x 2 - 2 v ớ i x ≠ ± 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)
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)`
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
\(\dfrac{20x\left(2-x\right)}{12x\left(x-2\right)^2}=\dfrac{5.4.x\left(2-x\right)}{3.4.x\left(2-x\right)^2}=\dfrac{5}{3\left(2-x\right)}\)
\(=\dfrac{-20x\left(x-2\right)}{12x\left(x-2\right)^2}=\dfrac{-5}{3\left(x-2\right)}\)