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\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
<=>\(\frac{1}{3}:2x=\frac{-21}{4}\)
<=>\(2x=\frac{1}{3}:\frac{-21}{4}\)
<=>\(2x=\frac{-4}{63}\)
<=>\(x=\frac{-2}{63}\)
|x - 1| = 2x
<=> x - 1 = 2x hoặc x - 1 = -2x
x - 1 - 2x = 0 x - 1 + 2x = 0
-x - 1 = 0 3x - 1 = 0
-x = 0 + 1 3x = 0 + 1
-x = 1 3x = 1
x = -1 x = 1/3
<=> x = -1 (ktm); x = 1/3 (tm)
=> x = 1/3
\(\left|x-1\right|=2x\)
\(TH1:x-1\ge0\Leftrightarrow x\ge1\)
Khi đó:\(\left|x-1\right|=2x\)
\(\Leftrightarrow x-1=2x\)
\(\Leftrightarrow x-2x=1\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(\text{loại vì không thỏa mãn điều kiện}\right)\)
\(TH2:x-1< 0\Leftrightarrow x< 1\)
Khi đó:\(\left|x-1\right|=2x\)
\(\Leftrightarrow-\left(x-1\right)=2x\)
\(\Leftrightarrow-x+1=2x\)
\(\Leftrightarrow-x-2x=-1\)
\(\Leftrightarrow-3x=-1\)
\(x=\frac{-1}{-3}=\frac{1}{3}\left(\text{nhận vì thỏa mãn điều kiện}\right)\)
Vậy \(S=\left\{\frac{1}{3}\right\}\)
GTTD cua mot so luon luon lon hon hoac bang 0
a)=>|4-0,5x|=0
=>A=-17 ( de A dat GTLN)
b) =>2|1/2.x-8|=0
=>x=16=>B=5 (de B dat GTLN)
c) de j the ?
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ròi tau làm choooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
1,
2x + 5 = 19
2x = 19 - 5
2x = 14
x = 14 : 2
x = 7
2,
( 1 - 2x ) + 5 = 10
( 1 - 2x ) = 5
2x = 1 - 5
2x = -4
x = -4 : 2
x = -2
1) 2x + 5 = 19
2x = 19 - 5
2x = 14
x = 14 : 2
x= 7
2) (1-2x) + 5 = 10
(1- 2x) = 10 - 5
(1-2x) = 5
2x = 1-5
2x = -4
x= -4 : 2
x = -2
1) 2x +5 =19
2x =19-5
2x =14
x =14:2
x =7
2) (1-2x)+5=10
(1-2x) =10-5
(1-2x) =5
2x =1-5
2x =-4
x =(-4):2
x = -2
k mik nha
1) 2x + 5 = 19
2x = 19 - 5
2x = 14
x = 7.
Vậy x = 7
2)( 1 - 2x ) + 5 = 10
1 - 2x = 10 - 5
1 - 2x = 5
2x = 1 - 5
2x = -4
x = -2
Vậy x = -2
1) \(2x+5=19\)
\(2x=19-5\)
\(2x=14\)
\(x=14\div2\)
\(x=7\)
2) \(\left(1-2x\right)+5=10\)
\(1-2x=10-5\)
\(1-2x=5\)
\(2x=1-5\)
\(2x=-4\)
\(x=\left(-4\right)\div2\)
\(x=-2\)
1) \(x\)\(=\)\(7\)
\(x\)\(=\)\(-2\)
Chúc bn hok tốt !!!
1)2x+5=19
2x =19-5
2x =14
x =14:2
x =7
Vậy x=7
2)(1-2x)+5=10
1-2x =10-5
1-2x =5
2x =1-5
2x =-4
x =-4:2
x =-2
Vậy x=-2