Tìm x biết: a) $\frac{8}{7}-\frac{1}{7}:(\frac{x}{3}-2)=-1$ b) $(|x+2|+\frac{1}{2}).(x^{2}-4)=0$
2 câu trả lời
`a) 8/7 - 1/7 : ( x/3 - 2 ) = - 1`
`⇒ 1/7 : ( x/3 - 2 ) = 8/7 - ( - 1 )`
`⇒ 1/7 : ( x/3 - 2 ) = 15/7`
`⇒ x/3 - 2 = 1/7 : 15/7`
`⇒ x/3 - 2 = 1/15`
`⇒ x/3 = 1/15 + 2`
`⇒ x/3 = 31/15`
`⇒ 15 . x = 31 . 3`
`⇒ 15 . x = 93`
`⇒ x = 93 : 15`
`⇒ x = 6,2`
Vậy `x = 6,2 .`
`b) ( | x + 2 | + 1/2 ) . ( x^2 - 4 ) = 0`
`⇒` \(\left[ \begin{array}{l}| x + 2 | + \frac{1}{2} = 0\\x^2 - 4 = 0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}| x + 2 | = \frac{- 1}{2} ( Loại )\\x^2 = 4\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x^2 = 2^2\\x^2 = ( - 2 )^2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 2\\x = - 2\end{array} \right.\)
Vậy `x ∈ { 2 ; - 2 }`
Đáp án:
Giải thích các bước giải:
`a) 8/7 - 1/7:(x/3 - 2) = -1`
`1/7:(x/3 - 2) = 8/7 - (-1)`
`1/7:(x/3-2) = 15/7`
`x/3-2 = 1/7 : 15/7 `
`x/3-2 = 1/15`
`x/3 = 1/15 + 2`
`x/3 = 31/15`
`x = 31:15*3`
`x = 31/5`
Vậy `x = 31/5`
`b) (|x+2|+1/2) *(x^2-4) = 0`
`<=>` $\left[\begin{matrix} |x+2|+\dfrac{1}{2} = 0\\ x^2-4=0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} |x+2|=\dfrac{-1}{2} \text{(loại)}\\ x^2=4\end{matrix}\right.$
`=> x^2 = 4`
`=> x^2 = 2^2` hoặc `(-2)^2`
`=> x = 2` hoặc `-2`
Vậy `x in {2;-2}`
