Đáp án:
Gửi tuss nhaaa~
Xin hn + 5sao + cam ơn!!!
@Mee210
Giải thích các bước giải:
`a, |x| = `1/5`
\(\left[ \begin{array}{x}x=`1/5`\\x=`-1/5`\end{array} \right.\)
`b, |x| = 0.37`
⇒ \(\left[ \begin{array}{x}x=0,37\\x=-0,37\end{array} \right.\)
`c) |x| = 0`
⇒ `x = 0`
`d) |x| = `1``2/3`
⇒\(\left[ \begin{array}{x}x=1`2/3`\\x=01`2/3`\end{array} \right.\)
Tìm x:
`a/` |`x- 1,7`| `= 2,3`
TH1: `x - 1.7 = 2,3`
`x = 2,3 + 1.7`
`x = 4`
TH2: `x - 1.7 =` `-2,3`
`x = ``-2,3 + 1.7`
`x =` `-3/5`
Vậy `x ∈ { 4;` `-3/5``}`
`b/` |`x+``3/4`| -`1/3``=0`
|`x`+`3/4`| = `1/3`
TH1: `x + ` `3/4` = `1/3`
`x = ` `1/3` - `3/4`
`x =` `-5/12`
TH2: `x + ` `3/4` = `-1/3`
`x = ` `-1/3` - `3/4`
`x =` `-13/12`
Vậy `x ∈ {` `-5/12`; `-13/12`}