Tìm x : 2.3^x .3^2 = 18
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DV
0
NT
3
1 tháng 5 2017
b) x + 2/5 = 2-3x/3
x + 2/5 = 2 -x
x +x = 2 - 2/5
2x= 8/5
x = 8/5 x 1/2
x = 4/5
1 tháng 5 2017
c) 2. 3^x . 3^2 = 18
3^x+2 = 18:2
3^ x+2 = 9
3^ x+2 = 3^2
=> x+2 = 2
=> x=0
VV
1
24 tháng 10 2015
(x-2)^20-(x-2)^18=0
(x-2)^18[(x-2)^2-1]=0
suy ra (x-2)^18=0;(x-2)^2-1=0
x-2=0 ;(x-2)^2=1 ;(x-2)^2=-1
x=2 ;x-2=1 ;x-2=-1
x=3 x=1
OH
0
6 tháng 11 2016
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HT
30 tháng 9 2015
a, 2.3x = 18
=> 3x = 9 = 32
=> x = 2
b, 2x + 3.2x + 4.2x = 32
=> 2x.(1+3+4) = 32
=> 2x . 8 = 32
=> 2x = 4 = 22
=> x = 2
c, 3x + 4x = 5x
Câu này mik chưa nghĩ ra
30 tháng 9 2015
a)1.3^x=18
=>3^x=18:2=9
=>3^x=3^2
=>x=2
b)=>2^x.(1+3+4)=32
=>2^x.8=32
=>2^x=4
=>2^x=2^2
=>x=2
13 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`
`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`
`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`
`=> x^2 . 11 = 2^4 . 11`
`=> x^2 . 11 - 2^4 . 11 = 0`
`=> 11 . (x^2 - 16) = 0`
`=> x^2 - 16 = 0`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = `\(\pm4\)
Vậy, `x \in`\(\left\{4;-4\right\}\)
_____
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)
`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)
`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)
`=>`\(\dfrac{23}{108}x=48-6^2\)
`=>`\(\dfrac{23}{108}x=48-36\)
`=>`\(\dfrac{23}{108}x=12\)
`=>`\(x=\dfrac{1296}{23}\)
Vậy, `x = `\(\dfrac{1296}{23}\)
13 tháng 7 2023
\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)
\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)
\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)
\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)
\(\Rightarrow x\in\varnothing\)
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)
\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)
2.3x .32 =18
3x+2 =9
3x+2 = 33 =>x+2 =3 =>x=1