Question

tìm x biết

a) \(\dfrac{2x-3}{3}+-\dfrac{3}{2}=\dfrac{5-2x}{6}-\dfrac{1}{2}\)

b) \(\left(3x-\dfrac{1}{2}\right).\left(3-\dfrac{1}{2}x\right).\left(1+\dfrac{3}{4}:x\right)=0\)

c) \(\left(3-\dfrac{9}{10}-|x+2|\right):\left(\dfrac{19}{20}-1-\dfrac{2}{3}\right)+\dfrac{4}{5}=1\)

Answer 1

Answer 2

a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)

<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)

<=>\(4x-17=0\)

<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)

Answer 3

b) =>\(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\3-\dfrac{1}{2}x=0\\1+\dfrac{3}{4}:x=0\end{matrix}\right.=>\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}x=3\\\dfrac{3}{4x}=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\x=6\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Answer 4