Tìm x biết a) |x+1/3|-5=7 b) 3(x-2)+2/5=4 17/07/2021 Bởi Remi Tìm x biết a) |x+1/3|-5=7 b) 3(x-2)+2/5=4
Đáp án + Giải thích các bước giải: a) `|x+1/3|-5=7` `=>|x+1/3|=12` `=>` \(\left[ \begin{array}{l}x+\dfrac13=12\\x+\dfrac13=-12\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\dfrac{35}{3}\\x=-\dfrac{37}{3}\end{array} \right.\) Vậy `x∈{35/3;-37/3}` b) `3(x-2)+2/5=4` `=>3(x-2)=18/5` `=>x-2=6/5` `=>x=16/5` Vậy `x=16/5` Bình luận
a) `|x+1/3| – 5 =7` `|x+1/3| = 7+5` `|x+1/3| = 12` +) `x+1/3 = 12` `x= 12 – 1/3` `x= 35/3` +) `x+1/3 = -12` `x = -12 -1/3` `x=-37/3` Vậy `x= 35/3` hoặc `x= -37/3` b) `3(x-2) + 2/5 =4` `3x – 6 +2/5 =4` `3x -28/5 =4` `3x = 4 + 28/5` `3x= 48/5` `x= 48/5 : 3` `x= 48/5 . 1/3` `x= 16/5` Vậy `x= 16/5` Bình luận
Đáp án + Giải thích các bước giải:
a)
`|x+1/3|-5=7`
`=>|x+1/3|=12`
`=>` \(\left[ \begin{array}{l}x+\dfrac13=12\\x+\dfrac13=-12\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{35}{3}\\x=-\dfrac{37}{3}\end{array} \right.\)
Vậy `x∈{35/3;-37/3}`
b)
`3(x-2)+2/5=4`
`=>3(x-2)=18/5`
`=>x-2=6/5`
`=>x=16/5`
Vậy `x=16/5`
a) `|x+1/3| – 5 =7`
`|x+1/3| = 7+5`
`|x+1/3| = 12`
+) `x+1/3 = 12`
`x= 12 – 1/3`
`x= 35/3`
+) `x+1/3 = -12`
`x = -12 -1/3`
`x=-37/3`
Vậy `x= 35/3` hoặc `x= -37/3`
b) `3(x-2) + 2/5 =4`
`3x – 6 +2/5 =4`
`3x -28/5 =4`
`3x = 4 + 28/5`
`3x= 48/5`
`x= 48/5 : 3`
`x= 48/5 . 1/3`
`x= 16/5`
Vậy `x= 16/5`