Question

tìm x

a, \(\dfrac{x}{12}\) - \(\dfrac{5}{6}\) = \(\dfrac{1}{12}\)

b, \(\dfrac{2}{3}\) - 1\(\dfrac{4}{15}\)x = \(\dfrac{-3}{5}\)

c, \(\dfrac{\left(-3\right)^x}{81}\) = -27

d, \(2^{x-1}\) = 16

e, \(\left(x-1\right)^2\) = 25

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

Answer 1

a) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{10}{12}\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\Rightarrow x=11\)

b) \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{10}{15}-\dfrac{19}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{-19}{15}x=\dfrac{-13}{15}\Rightarrow x=\dfrac{13}{19}\)

c) \(\dfrac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x=-2187\Rightarrow x=7\)

d) \(2^{x-1}=16\Rightarrow x-1=4\Rightarrow x=5\)

e) \(\left(x-1\right)^2=25\Rightarrow x-1=5\Rightarrow x=6\)

g) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{12}\\x+\dfrac{1}{2}=0\Rightarrow x=\dfrac{-1}{2}\end{matrix}\right.\)