tìm x,biết a) |4x+1|=1/3 b) |4x-6|= -2 c) |0,375x – 2,5|= 5/3 giúp mình với ,mình đang cần gấp :(( 22/08/2021 Bởi Remi tìm x,biết a) |4x+1|=1/3 b) |4x-6|= -2 c) |0,375x – 2,5|= 5/3 giúp mình với ,mình đang cần gấp :((
Đáp án: CHÚC BẠN HỌC TỐT!!! Giải thích các bước giải: $a, |4x+1|=\dfrac{1}{3}$ \(⇔\left[ \begin{array}{l}4x+1=\dfrac{1}{3}\\4x+1=-\dfrac{1}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}4x=-\dfrac{2}{3}\\4x=-\dfrac{4}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=-\dfrac{1}{3}\end{array} \right.\) $b, |4x-6|=-2$ Vì $|4x-6| \geq 0⇒PTVN$ $c, |0,375x-2,5|=\dfrac{5}{3}$ \(⇔\left[ \begin{array}{l}0,375x-2,5=\dfrac{5}{3}\\0,375x-2,5=-\dfrac{5}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}0,375x=\dfrac{25}{6}\\0,375x=\dfrac{5}{6}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{100}{9}\\x=\dfrac{20}{9}\end{array} \right.\) Bình luận
$a$) `|4x+1| = 1/3` `⇒` \(\left[ \begin{array}{l}4x+1=\dfrac{1}{3}\\4x+1=-\dfrac{1}{3}\end{array} \right.\) $⇒$ \(\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=-\dfrac{1}{3}\end{array} \right.\) Vậy `x` `∈` `{-1/6;-1/3}` $b$) `|4x-6| = -2` Vì : `|4x-6| ≥0` `∀` `x` `⇒` `x` `∈` `∅` Vậy `x` `∈` `∅` $c$) `|0,375 . x – 2,5| = 5/3` `⇔ |3/8 . x – 5/2| = 5/3` `⇒` \(\left[ \begin{array}{l}\dfrac{3}{8}x-\dfrac{5}{2}=\dfrac{5}{3}\\\dfrac{3}{8}x-\dfrac{5}{2}=-\dfrac{5}{3}\end{array} \right.\) $⇒$ \(\left[ \begin{array}{l}x=\dfrac{100}{9}\\x=\dfrac{20}{9}\end{array} \right.\) Vậy `x` `∈` `{{100}/9;{20}/9}` Bình luận
Đáp án:
CHÚC BẠN HỌC TỐT!!!
Giải thích các bước giải:
$a, |4x+1|=\dfrac{1}{3}$
\(⇔\left[ \begin{array}{l}4x+1=\dfrac{1}{3}\\4x+1=-\dfrac{1}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}4x=-\dfrac{2}{3}\\4x=-\dfrac{4}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=-\dfrac{1}{3}\end{array} \right.\)
$b, |4x-6|=-2$
Vì $|4x-6| \geq 0⇒PTVN$
$c, |0,375x-2,5|=\dfrac{5}{3}$
\(⇔\left[ \begin{array}{l}0,375x-2,5=\dfrac{5}{3}\\0,375x-2,5=-\dfrac{5}{3}\end{array} \right.\) \(⇔\left[ \begin{array}{l}0,375x=\dfrac{25}{6}\\0,375x=\dfrac{5}{6}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{100}{9}\\x=\dfrac{20}{9}\end{array} \right.\)
$a$) `|4x+1| = 1/3`
`⇒` \(\left[ \begin{array}{l}4x+1=\dfrac{1}{3}\\4x+1=-\dfrac{1}{3}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy `x` `∈` `{-1/6;-1/3}`
$b$) `|4x-6| = -2`
Vì : `|4x-6| ≥0` `∀` `x`
`⇒` `x` `∈` `∅`
Vậy `x` `∈` `∅`
$c$) `|0,375 . x – 2,5| = 5/3`
`⇔ |3/8 . x – 5/2| = 5/3`
`⇒` \(\left[ \begin{array}{l}\dfrac{3}{8}x-\dfrac{5}{2}=\dfrac{5}{3}\\\dfrac{3}{8}x-\dfrac{5}{2}=-\dfrac{5}{3}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{100}{9}\\x=\dfrac{20}{9}\end{array} \right.\)
Vậy `x` `∈` `{{100}/9;{20}/9}`