Bài 4:
a: \(\dfrac{x}{9}=-\dfrac{2}{5}\)
=>\(x=-9\cdot\dfrac{2}{5}=-\dfrac{18}{5}\)
b: ĐKXĐ: x<>1
\(\dfrac{0.5}{x-1}=\dfrac{2}{-0,2}\)
=>\(x-1=\dfrac{-0,2\cdot0,5}{2}=\dfrac{-0,1}{2}=-\dfrac{1}{20}\)
=>\(x=-\dfrac{1}{20}+1=\dfrac{19}{20}\left(nhận\right)\)
c: \(\left(2x-0,6\right)\left(\dfrac{3}{7}+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-0,6=0\\x+\dfrac{3}{7}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=0,6\\x=-\dfrac{3}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
d: \(\left(2x+2,5\right)^2=9\)
=>\(\left[{}\begin{matrix}2x+2,5=3\\2x+2,5=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=0,5\\2x=-5,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,25\\x=-2,75\end{matrix}\right.\)
e: \(17-\left(\dfrac{7}{2}-x\right)^2=1\)
=>\(\left(\dfrac{7}{2}-x\right)^2=17-1=16\)
=>\(\left(x-\dfrac{7}{2}\right)^2=16\)
=>\(\left[{}\begin{matrix}x-\dfrac{7}{2}=4\\x-\dfrac{7}{2}=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
f: \(x\cdot0,5+x\cdot1,5=-4\)
=>\(x\cdot\left(0,5+1,5\right)=-4\)
=>2x=-4
=>x=-2
g: \(-x\cdot\dfrac{2}{5}+\dfrac{7}{5}\cdot x=0,9\)
=>\(x\left(\dfrac{7}{5}-\dfrac{2}{5}\right)=0,9\)
=>\(x\cdot1=0,9\)
=>x=0,9
h: \(30\%\cdot x-x+\dfrac{1}{5}=0,6\)
=>\(x\left(0,3-1\right)=0,6-0,2=0,4\)
=>\(x\cdot\left(-0,7\right)=0,4\)
=>\(x=-\dfrac{4}{7}\)
Bài 3:
a) \(x-\dfrac{4}{7}=\dfrac{-3}{2}\) b) \(\dfrac{1}{7}+x=\dfrac{3}{14}\)
x = \(\dfrac{-3}{2}+\dfrac{4}{7}\) \(x=\dfrac{3}{14}-\dfrac{1}{7}\)
x = \(\dfrac{-21}{12}+\dfrac{8}{12}\) \(x=\dfrac{3}{14}-\dfrac{2}{14}\)
x = \(\dfrac{-13}{12}\) \(x=\dfrac{1}{14}\)
Vậy x = \(\dfrac{-13}{12}\) Vậy x = \(\dfrac{1}{14}\)
c) \(x:\dfrac{8}{5}=-5\) \(x.0,5=\dfrac{-1}{3}\)
x = -5 . \(\dfrac{8}{5}\) \(x.\dfrac{1}{2}=\dfrac{-1}{3}\)
x = \(\dfrac{-5}{1}.\dfrac{8}{5}\) x = \(\dfrac{-1}{3}:\dfrac{1}{2}\)
x = -8 x = \(\dfrac{-1}{3}.\dfrac{2}{1}\)
Vậy x = -8 x = \(\dfrac{-2}{3}\)
Vậy x = \(\dfrac{-2}{3}\)
e) \(\dfrac{2}{5}.x=0,4-\dfrac{12}{5}\) f) \(\dfrac{1}{3}+\dfrac{2}{3}:x=0,75\)
\(\dfrac{2}{5}.x=\dfrac{2}{5}-\dfrac{12}{5}\) \(\dfrac{1}{3}+\dfrac{2}{3}:x=\dfrac{3}{4}\)
\(\dfrac{2}{5}.x=\dfrac{-10}{5}\) \(\dfrac{2}{3}:x=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{2}{5}.x=-2\) \(\dfrac{2}{3}:x=\dfrac{9}{12}-\dfrac{4}{12}\)
\(x=-2:\dfrac{2}{5}\) \(\dfrac{2}{3}:x=\dfrac{5}{12}\)
\(x=\dfrac{-2}{1}.\dfrac{5}{2}\) \(x=\dfrac{2}{3}:\dfrac{5}{12}\)
\(x=-5\) \(x=\dfrac{2}{3}.\dfrac{12}{5}\)
Vậy x = -5 \(x=\dfrac{2}{1}.\dfrac{4}{5}\)
\(x=\dfrac{8}{5}\)
Vậy \(x=\dfrac{8}{5}\)
g) \(\dfrac{1}{4}+\dfrac{3}{5}:\left(2x-1\right)=1\)
\(\dfrac{3}{5}:\left(2x-1\right)=1-\dfrac{1}{4}\)
\(\dfrac{3}{5}:\left(2x-1\right)=\dfrac{3}{4}\)
\(2x-1=\dfrac{3}{5}:\dfrac{3}{4}\)
\(2x-1=\dfrac{3}{5}.\dfrac{4}{3}\)
\(2x-1=\dfrac{4}{5}\)
\(2x\) \(=\dfrac{4}{5}+1\)
\(2x\) \(=\dfrac{9}{5}\)
\(x\) \(=\dfrac{9}{5}:2\)
\(x\) \(=\dfrac{9}{5}.\dfrac{1}{2}\)
\(x\) \(=\dfrac{9}{10}\)
Vậy \(x=\dfrac{9}{10}\)