2 mũ x+1- 2 mũ x= 64
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1a.4,9,16,25,36,49,64,91,110,121,144,169,400,900,1600
b. 8,27,64,125,1000
C1111^2=1234321
2. x=6 ,x=3,x=0, x k thuộc n, x k thuộc n
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(x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
(2x - 1)3 = -64
=> (2x - 1)3 = -43
=> 2x - 1 = -4
=> 2x = -4 + 1
=> 2x = -3
=> x = -3/2
( x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
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a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
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a: \(3^x-2=2^7\)
\(\Leftrightarrow3^x=128+2=130\)(vô lý)
b: \(4^{x+1}=64\)
=>x+1=3
hay x=2
c: \(\left(5x+1\right)^2=1^{2016}=1\)
=>5x+1=1 hoặc 5x+1=-1
=>x=0 hoặc x=-2/5
d: \(2^{2\left(x-1\right)}=8\)
=>2(x-1)=3
=>x-1=3/2
hay x=5/2
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a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
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a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)
b) \(\left(4x-1\right)^2=16x^2-8x+1\)
c) \(\left(x+2\right)^2=x^2+4x+4\)
d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)
e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)
f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)
g) \(\left(x+1\right)^3=x^3+3x^2+3x+1\)
h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)
k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)
y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)
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\(m,x^3+48x=12x^2+64\)
\(x^3+48x-12x^2-64=0\)
\(\left(x-4\right)^3=0\)
\(x=4\)
\(n,x^3-3x^2+3x=1\)
\(x^3-3x^2+3x-1=0\)
\(\left(x-1\right)^3=0\)
\(x=1\)
\(\Leftrightarrow x^3+48x-12x^2-64=0\)0
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)-12x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)^3=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
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a) 5x = 125
5x = 53
=> x = 3
b) x3 = 64
x3 = 43
=> x = 4
c) ( x - 1 ) 2 = 25
( x - 1 ) 2 = 52
=> x - 1 = 5
=> x = 5 + 1
=> x = 6
d) 5x + 5x+2 = 130
5x . 1 + 5x . 52 = 130
5x . ( 1 + 52 ) = 130
5x . 26 = 130
5x = 130 : 26
5x = 5
=> x = 1
TL:
\(2^{x+1}-2^x=64\)
\(2^x\left(2-1\right)=64\)
\(2^x.1=64\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
Vậy x=6
HT!~!
2x+1 - 2x= 64
2x.21 - 2x = 28
2x (2-1) = 28
2x = 28
x = 8
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