Tìm x:
a) 5x = 125
b) 32x = 81
c) 52x-3 - 2.52 = 52.3
Nói trước x là số x dấu . là nhân
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a). 5x = 125
=>5x = 53
=> x = 3
b) 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
c). 52x-3 – 2.52 = 52 .3
=>52x: 53 = 52 .3 + 2.52
=>52x: 53 = 52 .5
=>52x = 52 .5.53
=>52x = 56
=> 2x = 6
=> x=3
a) 5x = 125
x = 125 : 5
x = 25
b) 32x = 81
x = 81 : 32
x = 81/ 32
c) 52x - 3 - 2.52 = 52.3
52x - 3 - 104 = 156
x - 3 - 104 = 156 : 52
x - 3 - 104 = 3
x - 3 = 3 + 104
x - 3 = 107
x = 107 + 3
x = 110
Ko chắc lắm , sai thì Sorry nhé .
b) 32x = 81
32x = 34
=> 2x = 4
=> x = 2
a). 52x-3 – 2.52 = 52 .3
52x: 53 = 52 .3 + 2.52
52x: 53 = 52 .5
52x = 52 .5.53
52x = 56
=> 2x = 6
=> x=3
a) \(8x+56:14=60\)
\(\Rightarrow8x+4=60\)
\(\Rightarrow8x=56\)
\(\Rightarrow x=\dfrac{56}{8}\)
\(\Rightarrow x=7\)
b) Mình làm rồi nhé !
c) \(41-2^{x+1}=9\)
\(\Rightarrow2^{x+1}=41-9\)
\(\Rightarrow2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
d) \(3^{2x-4}-x^0=8\)
\(\Rightarrow3^{2x-4}-1=8\)
\(\Rightarrow3^{2x-4}=9\)
\(\Rightarrow3^{2x-4}=3^2\)
\(\Rightarrow2x-4=2\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
g) \(65-4^{x+2}=2014^0\)
\(\Rightarrow65-4^{x+2}=1\)
\(\Rightarrow4^{x+2}=64\)
\(\Rightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=1\)
i) \(120+2\left(4x-17\right)=214\)
\(\Rightarrow2\left(4x-17\right)=214-120\)
\(\Rightarrow2\left(4x-17\right)=94\)
\(\Rightarrow4x-17=47\)
\(\Rightarrow4x=47+17\)
\(\Rightarrow4x=64\)
\(\Rightarrow x=16\)
a: \(8x+56:14=60\)
=>8x+4=60
=>8x=60-4=56
=>x=56/8=7
b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)
=>2x-3=3
=>2x=6
=>x=3
c: \(41-2^{x+1}=9\)
=>\(2^{x+1}=41-9=32\)
=>x+1=5
=>x=4
d: \(3^{2x-4}-x^0=8\)
=>\(3^{2x-4}-1=8\)
=>\(3^{2x-4}=8+1=9\)
=>2x-4=2
=>2x=6
=>x=3
g: \(65-4^{x+2}=2014^0\)
=>\(65-4^{x+2}=1\)
=>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
i: 120+2(4x-17)=214
=>2(4x-17)=214-120=94
=>4x-17=94/2=47
=>4x=64
=>\(x=\dfrac{64}{4}=16\)
`a)(x-6)^2-(x+6)^2=12`
`<=>(x-6-x-6)(x-6+x+6)=12`
`<=>-12.2x=12`
`<=>2x=-1`
`<=>x=-1/2`
Vậy `x=-1/2`
`b)36x^2-12x+1=81`
`<=>(6x-1)^2=81`
`<=>(6x-1-9)(6x-1+9)=0`
`<=>(6x-10)(6x+8)=0`
`<=>(3x-5)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2-6x+2x-12=0`
`<=>x(x-6)+2(x-6)=0`
`<=>(x-6)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2-6x+x-6=0`
`<=>x(x-6)+x-6=0`
`<=>(x-6)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
a, 2.(x – 5)+7 = 77
<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40
b, x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14
<=> x - 1 3 - 3 + 2 4 = 14
<=> x - 1 3 = 14 + 3 - 16 = 1
<=> x – 1 = 1 <=> x = 2
c, 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1
Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A = 2 + 2 2 + 2 3 + . . . + 2 2017
=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )
=> A = 2 2017 - 1
Ta có: 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 => 2 2017 - 1 = 2 x - 1 - 1 => x = 2018
d, 5 2 x - 3 - 2 . 5 2 = 5 2 . 3
<=> 5 2 x - 3 = 5 2 . 3 + 5 2 . 2
<=> 5 2 x - 3 = 5 2 . ( 3 + 2 )
<=> 5 2 x - 3 = 5 3
<=> 2x – 3 = 3 => x = 3
a)5x=125=53
=>x=3
b)32x=81=34
=>2x=4
x=4/2
x=2
c)52x-3-2*52=52*3
52x-3=52*3+2*52=52(3+2)=52*5
52x-3=53
=>2x-3=3
2x=3+3
2x=6
x=6/2
x=3
lần sau nếu ghi dẫu mũ như vậy thì bạn nhớ ngoặc vào
chịu mới lớp 4 mà !!!
hỏi làm chi