tìm x
(x-1)^2=1/4 3/2 *x-1=x-1/2
5/6-3x=-2/3+1/2 *x
mong được giúp đỡ mình sẽ cho 5 sao nếu chi tiết
tìm x
(x-1)^2=1/4 3/2 *x-1=x-1/2
5/6-3x=-2/3+1/2 *x
mong được giúp đỡ mình sẽ cho 5 sao nếu chi tiết
`a) (x – 1)^2 = 1/4`
$\text { ⇒ \(\left[ \begin{array}{l}x – 1 = \frac{1}{2} \\x – 1 = \frac{-1}{2}\end{array} \right.\) }$
$\text { ⇒ \(\left[ \begin{array}{l}x = \frac{1}{2} + 1 \\x = \frac{-1}{2} + 1\end{array} \right.\) }$
$\text { ⇒ \(\left[ \begin{array}{l}x = \frac{1}{2} + \frac{2}{2} \\x = \frac{-1}{2} + \frac{2}{2}\end{array} \right.\) }$
$\text { ⇒ \(\left[ \begin{array}{l}x = \frac{3}{2} \\x = \frac{1}{2} \end{array} \right.\) }$
$\text { Vậy x ∈ { $\frac{3}{2}$ ; $\frac{1}{2}$ } }$
`b) 3/2 . x – 1 = x – 1/2`
`⇒ 3/2 . x – x = – 1/2 + 1` (áp dụng quy tắc chuyển vế)
`⇒ ( 3/2 – 1) . x = – 1/2 + 2/2`
`⇒ ( 3/2 – 2/2 ) . x = 1/2`
`⇒ 1/2 . x = 1/2`
`⇒ x = 1/2 : 1/2`
`⇒ x = 1`
$\text { Vậy x = 1 }$
`c) 5/6 – 3x = (-2)/3 + 1/2 . x`
`⇒ 5/6 + 2/3 = 1/2 . x + 3x` (áp dụng quy tắc chuyển vế)
`⇒ 5/6 + 4/6 = x . ( 1/2 + 3 )`
`⇒ 9/6 = x . ( 1/2 + 6/2 )`
`⇒ 9/6 = x . 7/2`
`⇒ x = 9/6 : 7/2`
`⇒ x = 9/6 . 2/7`
`⇒ x = 3/7`
$\text { Vậy x = $\frac{3}{7}$ }$
1, (x – 1)² = $\frac{1}{4}$
(x – 1)² = $\frac{1}{2}$²
⇒ \(\left[ \begin{array}{l}x-1=\frac{1}{2}\\x-1=\frac{-1}{2} \end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\frac{1}{2}+1\\x=\frac{-1}{2}+1\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\frac{1+2}{2}=\frac{3}{2}\\x=\frac{-1+2}{2}=\frac{1}{2}\end{array} \right.\)
2, $\frac{3}{2}$ x – 1 = x – $\frac{1}{2}$
$\frac{3}{2}$ x – x = $\frac{-1}{2}$ + 1
($\frac{3}{2}$ – 1) x = $\frac{-1+2}{2}$
($\frac{3-2}{2}$) x = $\frac{1}{2}$
$\frac{1}{2}$ x = $\frac{1}{2}$
x = $\frac{1}{2}$ : $\frac{1}{2}$
x = 1
3, $\frac{5}{6}$ – 3x = $\frac{-2}{3}$ + $\frac{1}{2}$ x
-3x – $\frac{1}{2}$ x = $\frac{-2}{3}$ – $\frac{5}{6}$
(-3 – $\frac{1}{2}$) x = $\frac{-4-5}{6}$
($\frac{-6-1}{2}$) x = $\frac{-9}{6}$
$\frac{-7}{2}$ x = $\frac{-9}{6}$
x = $\frac{-9}{6}$ : $\frac{-7}{2}$
x = $\frac{-9}{6}$ . $\frac{-2}{7}$
x = $\frac{3}{7}$
Chúc bạn học tốt ^^