Question

 

đề bài là tìm x

Answer 1

`c) sqrt{x^2} = 5/6`

`ĐK : x >0`

`<=> |x| =5/6`

`<=>`\(\left[{}\begin{matrix}x=\dfrac{5}{6}\left(tm\right)\\x=-\dfrac{5}{6}\left(ktm\right)\end{matrix}\right.\)

Vậy `x=5/6`

`d) sqrt{ (x+1)^2 } = 2/3`

`<=> |x+1|=2/3`

`<=> [(x+1=2/3),(x+1/=-2/3):}`

`<=> [(x=2/3-1=-1/3),(x=-2/3-1=-5/3):}`

Vậy ` in {-1/3;-5/3}`

Answer 2

c: =>|x|=5/6

mà x>0

nên x=5/6

d: =>|x+1|=2/3

=>x+1=2/3 hoặc x+1=-2/3

=>x=-1/3 hoặc x=-5/3