\(a,\left(-\frac{1}{2}-0,6\right).x=\left(-3\right)^2+\frac{1}{2}\)
=> \(-\frac{11}{10}.x=9+\frac{1}{2}\)
=> \(-\frac{11}{10}.x=\frac{19}{2}\)
=> \(x=\frac{19}{2}:-\frac{11}{10}=-\frac{95}{11}\)
Vậy \(x\in\left\{-\frac{95}{11}\right\}\)
\(b,\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-2\right)x=0,5\)
=> \(\frac{1}{3}-\frac{1}{3}x=0,5\)
=> \(\frac{1}{3}x=\frac{1}{3}-0,5=-\frac{1}{6}\)
=> \(x=-\frac{1}{6}:\frac{1}{3}=-\frac{1}{2}\)
Vậy \(x\in\left\{-\frac{1}{2}\right\}\)
\(c,\left|x\right|+0,325=2\)
=> \(\left|x\right|=2-0,325=1,675=\frac{67}{40}\)
=> \(\left[{}\begin{matrix}x=\frac{67}{40}\\x=-\frac{67}{40}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{67}{40};-\frac{67}{40}\right\}\)
\(d,\left|x+0,25\right|=6,25\)
=> \(\left[{}\begin{matrix}x+0,25=6,25\\x+0,25=-6,25\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=6,25-0,25=6\\x=-6,25-0,25-6.5=-\frac{13}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{6;-\frac{13}{2}\right\}\)
a) \(\left(-\frac{1}{2}-0,6\right).x=\left(-3\right)^2+\frac{1}{2}\)
=> \(-\frac{11}{10}.x=9+\frac{1}{2}\)
=> \(-\frac{11}{10}.x=\frac{19}{2}\)
=> \(x=\frac{19}{2}:\left(-\frac{11}{10}\right)\)
=> \(x=-\frac{95}{11}\)
Vậy \(x=-\frac{95}{11}\).
b) \(\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-2\right).x=0,5\)
=> \(\frac{1}{3}-\frac{1}{3}.x=\frac{1}{2}\)
=> \(\frac{1}{3}.x=\frac{1}{3}-\frac{1}{2}\)
=> \(\frac{1}{3}.x=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{6}\right):\frac{1}{3}\)
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\).
c) |x| + 0, 325 = 2
|x| = 2 - 0, 325
|x| = 1, 675
|x| = \(\frac{67}{40}\)
=> \(\left\{{}\begin{matrix}x=\frac{67}{40}\\x=-\frac{67}{40}\end{matrix}\right.\)
Vậy x ∈ \(\left\{\frac{67}{40};-\frac{67}{40}\right\}\).
d) |x + 0, 25| = 6, 25
TH1: x + 0, 25 = 6, 25
x = 6, 25 - 0, 25
x = 6
TH2: x + 0, 25 = -6, 25
x = (-6, 25) - 0, 25
x = -6, 5
Vậy x ∈ \(\left\{6;-6,5\right\}\).
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