Tìm x:
a) (2x - 3)(6 - 2x) = 0
b) \(5\dfrac{4}{7}:x=13\)
c) 2x - \(\dfrac{3}{7}\) = \(6\dfrac{2}{7}\)
d) \(\dfrac{x}{5}\) + \(\dfrac{1}{2}\) = \(\dfrac{6}{10}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
f) \(\dfrac{x-12}{4}=\dfrac{1}{2}\)
g) \(2\dfrac{1}{4}\).\(\left(x-7\dfrac{1}{3}\right)=1,5\)
h) \(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
i) \(\dfrac{2}{3}\left(x-25\%\right)=\dfrac{1}{6}\)
k) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)
⇒\(x-12=2\)
\(x=2+12\)
x = 14
g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)
\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\)
\(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)
\(x-\dfrac{22}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{22}{3}\)
\(x=8\)
h)(4,5 - 2x ) .1\(\dfrac{4}{5}\) = \(\dfrac{11}{4}\)
(4,5 - 2x) . 1,8 = 2,75
(4,5 - 2x) = 2,75 . 1,8
(4,5 - 2x) = 4,95
2x = 4,95
x = 4,95 : 2
x = 2,475
i)\(\dfrac{2}{3}.\left(x-25\%\right)=\dfrac{1}{6}\)
\(x-0,25=\dfrac{1}{6}:\dfrac{2}{3}\)
\(x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(x=\dfrac{1}{2}\)
k)\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\dfrac{3}{2}x-x=-\dfrac{3}{4}+\dfrac{3}{2}\)
\(\dfrac{1}{2}x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{1}{2}\)
\(x=\dfrac{3}{2}\)
