Question

Bài 2. Tìm số hữu tỉ x, biết:

a) 4x + \(\dfrac{1}{3}\) = \(\dfrac{-3}{2}\)                     b) \(\dfrac{-3}{4}\)x - \(\dfrac{3}{5}\) = \(\dfrac{-11}{10}\)

c) \(\left(\dfrac{1}{3}\right)^{3x}\) = \(\dfrac{1}{729}\)

Answer 1

a: \(4x+\dfrac{1}{3}=-\dfrac{3}{2}\)

=>\(4x=-\dfrac{3}{2}-\dfrac{1}{3}=-\dfrac{9}{6}-\dfrac{2}{6}=-\dfrac{11}{6}\)

=>\(x=-\dfrac{11}{6}:4=-\dfrac{11}{24}\)

b: \(-\dfrac{3}{4}x-\dfrac{3}{5}=-\dfrac{11}{10}\)

=>\(-\dfrac{3}{4}x=-\dfrac{11}{10}+\dfrac{3}{5}=-\dfrac{11}{10}+\dfrac{6}{10}=-\dfrac{5}{10}=-\dfrac{1}{2}\)

=>\(x=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\)

c: \(\left(\dfrac{1}{3}\right)^{3x}=\dfrac{1}{729}\)

=>\(\left(\dfrac{1}{3}\right)^{3x}=\left(\dfrac{1}{3}\right)^6\)

=>3x=6

=>\(x=\dfrac{6}{3}=2\)

Answer 2

\(a,4x+\dfrac{1}{3}=\dfrac{-3}{2}\)
\(4x=\dfrac{-3}{2}-\dfrac{1}{3}\)
\(4x=\dfrac{-9}{6}-\dfrac{2}{6}\)
\(4x=\dfrac{-11}{6}\)
\(x=\dfrac{-11}{6}:4\)
\(x=\dfrac{-11}{24}\)
\(b,\dfrac{-3}{4}x-\dfrac{3}{5}=\dfrac{-11}{10}\)
\(\dfrac{-3}{4}x=\dfrac{-11}{10}+\dfrac{3}{5}\)
\(\dfrac{-3}{4}x=\dfrac{-11}{10}+\dfrac{6}{10}\)
\(\dfrac{-3}{4}x=\dfrac{-5}{10}\)
\(\dfrac{-3}{4}x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{-3}{4}\)
\(x=\dfrac{-1}{2}.\dfrac{-4}{3}\)
\(x=\dfrac{2}{3}\)