\(\dfrac{x-2}{x+2}-\dfrac{x-2}{x+2}=\dfrac{-26}{x^2-4}\)(1)
ĐKXĐ: x≠+-2
(1)⇒(x-2)(x+2)-(x-2)(x+2)=-26
⇔\(x^2\)+2x-2x-4-x-\(x^2\)+2x-2x+4=-26
⇔0=-26
vậy S=∅
\(\dfrac{x-2}{x+2}-\dfrac{x-2}{x+2}=\dfrac{-26}{x^2-4}\)(1)
ĐKXĐ: x≠+-2
(1)⇒(x-2)(x+2)-(x-2)(x+2)=-26
⇔\(x^2\)+2x-2x-4-x-\(x^2\)+2x-2x+4=-26
⇔0=-26
vậy S=∅
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-26}{x^2-4}\)
Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số
1)\(\dfrac{x+2}{3}>\dfrac{x}{2}+\dfrac{1}{6}\)
2) 2x(6x-1)>(3x-2)(4x+3)
3) \(\dfrac{2\left(x+1\right)}{3}\)-2≥\(\dfrac{x-2}{2}\)
4)2-5x≤17
5) \(\dfrac{x+2}{5}-\dfrac{x-2}{3}\) <2
6) \(\dfrac{x+2}{3}< \dfrac{3-2x}{5}\)
7)\(\dfrac{4\left(x-1\right)}{3}-\dfrac{2-x}{15}\) <\(\dfrac{10x-3}{5}\)
8) 2x-\(\dfrac{x+2}{3}\) <\(\dfrac{3\left(x-2\right)}{2}\)+5-x
9) 2x-3(x+1)>6x+3(x-5)
10) \(\dfrac{2x+3}{7}\) >\(\dfrac{x-5}{4}\)
giúp mik giải bài này vs mik đag cần gấp mik c.ơn
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
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
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
giải các phương trình sau
1, \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
2, \(\dfrac{3}{2+x}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
3, \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
giải các phương trình:
\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\dfrac{x+2}{2020}+\dfrac{x+4}{2018}=\dfrac{x+6}{2016}+\dfrac{x+8}{2014}\)
Giúp tớ với, thầy réo tớ kinh lắm rồi!
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
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) \(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{-16}{x^2-4}\)