tìm x biết a) ( 3/2x+10 ):2/3=6 b) 1/3x-0,5x=0,75 c) |x+1|=2 ².(-1 1/4)+15 Hộ với ạ 15/08/2021 Bởi Alice tìm x biết a) ( 3/2x+10 ):2/3=6 b) 1/3x-0,5x=0,75 c) |x+1|=2 ².(-1 1/4)+15 Hộ với ạ
`a)( 3/2x+10 ):2/3=6``=> 3/2x + 10 = 6 . 2/3``=> 3/2x + 10 = 4``=> 3/2x = 4-10``=>3/2x = -6``=> x = -6 : 3/2``=> x = -4`Vậy `x=-4` ` b) 1/3x-0,5x=0,75`` => x . (1/3-0,5) = 0,75``=> x. (-1)/6 = 0,75``=> x = 0,75 : (-1)/6``=> x= (-9)/2`Vậy `x=(-9)/2` `c) |x+1|=2^2.(-1 1/4)+15`` => |x+1| = 4.(-5)/4 +15``=> |x+1| = -5 + 15``=>|x+1| = 10``=> x+1=10` hoặc `x+1=-10``+) x+1=10 =>x=9``+) x+1=-10=>x=-11`Vậy `x\in{9;-11}` Bình luận
Đáp án: `a) ( 3/2x+10 ):2/3=6` `3/2 x + 10 = 6 . 2/3` `3/2 x + 10 = 4` `3/2 x = 4 – 10` `3/2 x = -6` `x = -6 : 3/2` `x = -4` Vậy `x = -4` b) `1/3 x-0,5x=0,75` `(1/3 – 0,5)x = 0,75` `-1/6x = 3/4` `x = 3/4 : -1/6` `x= -9/2` Vậy `x = -9/2` c) `|x+1|=2 ².(-1 1/4)+15` `|x + 1| = 4 . -5/4 + 15` `|x + 1| = -5 + 15` `|x + 1| = 10` `⇒` \(\left[ \begin{array}{l}x + 1 = 10\\x + 1 = -10\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x = 9\\x = -11\end{array} \right.\) Vậy `x ∈ {9 ; -11}` Bình luận
`a)( 3/2x+10 ):2/3=6`
`=> 3/2x + 10 = 6 . 2/3`
`=> 3/2x + 10 = 4`
`=> 3/2x = 4-10`
`=>3/2x = -6`
`=> x = -6 : 3/2`
`=> x = -4`
Vậy `x=-4`
` b) 1/3x-0,5x=0,75`
` => x . (1/3-0,5) = 0,75`
`=> x. (-1)/6 = 0,75`
`=> x = 0,75 : (-1)/6`
`=> x= (-9)/2`
Vậy `x=(-9)/2`
`c) |x+1|=2^2.(-1 1/4)+15`
` => |x+1| = 4.(-5)/4 +15`
`=> |x+1| = -5 + 15`
`=>|x+1| = 10`
`=> x+1=10` hoặc `x+1=-10`
`+) x+1=10 =>x=9`
`+) x+1=-10=>x=-11`
Vậy `x\in{9;-11}`
Đáp án:
`a) ( 3/2x+10 ):2/3=6`
`3/2 x + 10 = 6 . 2/3`
`3/2 x + 10 = 4`
`3/2 x = 4 – 10`
`3/2 x = -6`
`x = -6 : 3/2`
`x = -4`
Vậy `x = -4`
b) `1/3 x-0,5x=0,75`
`(1/3 – 0,5)x = 0,75`
`-1/6x = 3/4`
`x = 3/4 : -1/6`
`x= -9/2`
Vậy `x = -9/2`
c) `|x+1|=2 ².(-1 1/4)+15`
`|x + 1| = 4 . -5/4 + 15`
`|x + 1| = -5 + 15`
`|x + 1| = 10`
`⇒` \(\left[ \begin{array}{l}x + 1 = 10\\x + 1 = -10\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x = 9\\x = -11\end{array} \right.\)
Vậy `x ∈ {9 ; -11}`