=>(2x-3)(2x-1)>0
TH1: 2x-3>0 và 2x-1>0
=>x>3/2 và x>1/2
=>x>3/2
TH2: 2x-3<0 và 2x-1<0
=>x<3/2 và x<1/2
=>x<1/2
Vậy: x>3/2 hoặc x<1/2
2. Tìm x
a. \(\dfrac{4}{5}-3.\left|x\right|=\dfrac{1}{5}\) b. \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
c. (2x-8)(10-5x)=0 d. \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
a, \(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
b, \(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
tìm x \(\in\) Q biết rằng
\(\dfrac{11}{12}\) - ( \(\dfrac{2}{5}\) + x ) = \(\dfrac{2}{3}\)
2x \(\times\) ( x - \(\dfrac{1}{7}\) ) = 0
\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
a)\(2\cdot\text{|}3-2x\text{|}+\dfrac{1}{2}=\dfrac{5}{2}\) b)\(x^2\cdot\left(2^x-6\right)-2x^3=0\)
\(\dfrac{1}{3}\)x(2x-\(\dfrac{1}{3}\))\(^2\)-\(\dfrac{25}{27}\)=0
a\(\dfrac{-3}{4}x+1=\dfrac{5}{6}\)
b \(\left(2x-3\right).\left(x-5\right)=0\)
c \(\dfrac{1}{2}-|x+1|=0,25\)
Tìm x, biết:
\(\dfrac{1}{2}x+\dfrac{4}{5}=2x-\dfrac{8}{5}\)
\(\sqrt{x}=5\) (x ≥ 0)
x2 = 3
72x+72x+3=344 (5-x)(9x2-4)=0 \(\left|2-2x\right|\)-3,75=(-0,5)2
\(\sqrt{x-1}\)+\(\dfrac{2}{3}\)=1 (\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)x)2=2\(\dfrac{1}{4}\) giúp mình với!!!! cảm ơn nhìu=))❤
Bài 2: Tìm x,y,z biết:
a)\(\left(x-1\right)\)\(:\)\(\dfrac{2}{3}\)=\(\dfrac{-2}{5}\)
b) \(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{3}=0\)
c) \(\left|4x+2\right|=\left|6+2x\right|\)
