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5.2x-1 = 40
=>2x-1 = 8
=>2x-1 = 23
=>x - 1 = 3
=>x = 4
3x + 37 = 118
=>3x = 81
=>3x = 34
=>x = 4
5 . 2x-1 = 40
- > 2x-1 = 40 : 5 = 8
2x-1 = 23
- > x - 1 = 3 - > x = 4
3x + 37 = 118
3x = 118 - 37 = 81
3x = 34 - > x = 4
S = 1 + 9 + 92 + ... + 92017
9S = 9 + 92 + 93 + ... + 92018
- > 9S - S = 8S = ( 9 + 92 + ... + 92018 ) - ( 1 + 9 + ... + 92017 )
8S = 92018 - 1
= > S = ( 92018 - 1 ) : 8
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a) 120 - 5 . ( x + 2 ) = 45
5 . (x + 2) = 120 - 45
5 . (x + 2) = 75
x + 2 = 75 : 5
x + 2 = 15
x = 17
b) ( 2.x - 3 )2 = 49
( 2.x - 3 )2 = 72
( 2.x - 3 ) = 7
2x = 10
x = 5
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a, 9x-1= 9 ( vì 2 bên cùng có cơ so là 9 nên ta bớt 9 đi )
x-1 = 1
x = 1+1
x = 2
Vậy x = 2
b, 2x : 25 = 1
2x = 1 x 25
2x = 25 ( cũng có cơ so 2 nen ta bot 2 di )
x = 5
Vậy x = 5
k mk nha bạn
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\(3^{x+1}=9^x\)
\(3^{x+1}=\left(3^2\right)^x\)
\(3^{x+1}=3^{2x}\)
\(→x+1=2x\)
\(1=2x-x\)
\(1=x\)
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
3x+1 =9x
=>3x+1 =(32)x
=>3x+1 =32x
=>x+1 =2x
=>1=2x-x
=>x=1
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\(0\div x=0\).............................
\(\Rightarrow x\inℝ\)....................
Hk otots............................
Theo bai ta co: 9^x-1=9^1(vi9=9^1)
vi tim so mu nen ta co: x-1=1
x =1+1
x =2
Vay 9^2-1=9 ==> x=2