Question

Tìm x biết

a) \(\left|x+2,9\right|-1,4=0,3\)

b) \(x^2-\frac{5}{14}x=0\)

Answer 1

a, \(\left|x+2,9\right|-1,4=0,3\)

\(\left|x+2,9\right|=0,3+1,4\)

\(\left|x+2,9\right|=1,7\)

TH 1 :

=> x + 2,9 = 1,7

x = 1,7 - 2,9

x = -1,2

TH 2:

=> x + 2,9 = -1,7

x = -1,7 + 2,9

x = 4,6

Answer 2

b, \(x^2-\frac{5}{14}x=0\)

\(x^2.1-\frac{5}{14}x=0\)

\(x\left(1-\frac{5}{14}\right)x=0\)

\(x^2.\frac{9}{14}=0\)

\(x^2=0:\frac{9}{14}\)

\(x^2=0\)

=> x = 0

Answer 3

a) \(\left|x+2,9\right|-1,4=0,3\)

⇒ \(\left|x+2,9\right|=0,3+1,4\)

⇒ \(\left|x+2,9\right|=1,7\)

⇒ \(\left[{}\begin{matrix}x+2,9=1,7\\x+2,9=-1,7\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=1,7-2,9\\x=\left(-1,7\right)-2,9\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-1,2\\x=-4,6\end{matrix}\right.\)

Vậy \(x\in\left\{-1,2;-4,6\right\}.\)

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