`[x-2]/[x+2] > 1`
`<=>[x-2]/[x+2]-1 > 0`
`<=>[x-2-x-2]/[x+2] > 0`
`<=>[-4]/[x+2] > 0`
`=>x+2 < 0`
`<=>x < -2`
x-2/x+2 > 1
=> x-2>x+2
=> -2>2 ( vô lý)
=> Không có bất kì giá trị nào của x
Rút gọn
a)\(\dfrac{x}{x+1}+\dfrac{1}{x-1}-\dfrac{2x}{1-x^2}\)
b)\(\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}\)
c)\(\dfrac{2x^2-3x-9}{x^2-9}-\dfrac{x}{x+3}-\dfrac{x+3}{3-x}\)
d)\(\dfrac{x+3}{x-2}+\dfrac{x+2}{1-x}-\dfrac{4x-x^2}{x^2-3x+2}\)
giúp mik vs
cảm ơn <3
Rút gọn
a) \((\dfrac{2x^2+3x}{x^3+1}+\dfrac{1}{x^2-x+1}).\dfrac{x^2-x+1}{x}\)
b) \(\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right):\left(\dfrac{x+1}{x-2}-\dfrac{x+2}{x-1}\right)\)
c) \(\left(\dfrac{1}{x}+\dfrac{x}{x+1}\right).\dfrac{x^2+x}{x}\)
Thu gọn
a) \(\dfrac{1}{x-1}+\dfrac{2x}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
b) \(\dfrac{x+2}{x-3}=\dfrac{x^2+3x}{x^2-9}\)
c) \(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{-16}{x^2-4}\)
Giải các phương trình sau
d) \(\dfrac{1}{x-2}\)-\(\dfrac{6}{x+3}\)=\(\dfrac{5}{6-x^2-x}\)
e) \(\dfrac{2}{x+2}\)-\(\dfrac{2x^2+16}{x^3+8}\)=\(\dfrac{5}{x^2-2x+4}\)
f) \(\dfrac{x+1}{x^2+x+1}\)-\(\dfrac{x-1}{x^2-x+1}\)=\(\dfrac{2\left(x+2\right)^2}{x^6-1}\)
m)(\(\dfrac{2x}{x^2-1}\)+\(\dfrac{x-1}{2x+2}\)):\(\dfrac{x+1}{2x}\)+\(\dfrac{3}{1-x}\)
p)(\(\dfrac{2+x}{2-x}\)+\(\dfrac{4x^2}{x^2-4}\)-\(\dfrac{2-x}{2+x}\)):\(\dfrac{x^2-3x}{2x^2-x^3}\)
Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
Thực hiên phép tính
a)\(\dfrac{x^2+2}{x^3+1}\)-\(\dfrac{1}{x+1}\)
b)\(\dfrac{x}{x^2-2x}\)-\(\dfrac{x^2+4x}{x^3-4x}\)-\(\dfrac{2}{x^2+2x}\)
c)\(\dfrac{1}{2-2x}\)-\(\dfrac{3}{2+2x}\)+\(\dfrac{2x}{x^2-1}\)
Thực hiện phép tính
a) \(^{\dfrac{x^2+2}{x^3-1}}\) +\(\dfrac{2}{x^2+x+1}\) +\(\dfrac{1}{1-x}\)
b) \(\dfrac{1}{x+2}\) +\(\dfrac{3}{x^2-4}\) +\(\dfrac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)
c)\(\dfrac{1}{x-y}\) -\(\dfrac{3xy}{x^3-y^3}\) +\(\dfrac{x-y}{x^2+xy+y^2}\)
d) \(\dfrac{1}{a-b}\) +\(\dfrac{1}{a+b}\) +\(\dfrac{2a}{a^2+b^2}\) +\(\dfrac{4a^3}{a^4+b^4}\)
e) \(\dfrac{1}{a^2-a}\) + \(\dfrac{1}{a^2-3a+2}\) +\(\dfrac{1}{a^2-5a+6}\) +\(\dfrac{1}{a^2-7a+12}\)
\(\dfrac{1}{x}\) - \(\dfrac{2}{x+1}\) = \(\dfrac{3}{x^2+x}\)
\(\dfrac{1}{x2-3}\) - \(\dfrac{3}{x\left(2x-3\right)}\) = \(\dfrac{5}{x}\)
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
GIúp mình với ạ
Rút gọn:
a) A= \(\dfrac{x+y}{x-y}-\dfrac{x}{x+y}+\dfrac{2y^2}{x^2-y^2}\)
b) B= \(\dfrac{x}{x-2}-\dfrac{10}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-1}{x+3}\)
c) C= \(\dfrac{1}{x-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x^3-1}\)
