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a, \(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
c, \(x^{10}=1^x\)
\(\Rightarrow x\in\left\{1;0\right\}\)
\(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32=2^4\)
\(\Rightarrow x=4\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
Phần này mk ko bt làm đâu
\(x^{10}=1^x\)
\(\Rightarrow\)\(x^{10}=1\)
\(\Rightarrow x=1\)
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a) \(2^x=32\)=>\(2^x=2^5\)=>\(x=5\)
b) \(64\times4^x=4^5\)
\(4^3.4^x=4^5\)
\(4^{3+x}=4^5\)
=>\(3+x=5\)
=> x=2
c)\(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
=>\(2^x=2^5\)
=>\(x=5\)
a, 2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
b, 64 x 4 mũ x = 4 mũ 5
64 x 4 mũ x = 1024
4 mũ x = 1024 : 64
4 mũ x = 16
4 mũ x = 4 mũ 2
x = 2
c, 2 mũ x - 15 = 17
2 mũ x = 17 + 15
2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
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15 + ( x : 5 - 1 ) = 24
15 + ( x : 5 - 1 ) = 16
x : 5 - 1 = 16 - 15
x : 5 - 1 = 1
x : 5 = 1 + 1
x : 5 = 2
x = 10
Vậy x = 10
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\(x^{200}=x\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
\(x^{100}=1\)
\(\Rightarrow x=1\)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow2x-15=2x-15\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) \(\left(-175\right)-123+3+175+\left(-3\right)\)
\(=\left[\left(-175\right)+175\right]+\left[3+\left(-3\right)\right]-123\)
\(=0+0-123\)
\(=-123\)
b) \(17.85+15.17-150\)
\(=17.\left(85+15\right)-150\)
\(=17.100-150\)
\(=1700-150\)
\(=1550\)
c) \(176:\left(4.5^2-3.2^2\right)\)
\(=176:\left(4.25-3.4\right)\)
\(=176:\left(100-12\right)\)
\(=176:88\)
\(=2\)
2)
a) \(8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=8+12\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
b) \(20+2^3.x=5^2.4\)
\(\Rightarrow20+8.x=25.4\)
\(\Rightarrow20+8.x=100\)
\(\Rightarrow8.x=100-20\)
\(\Rightarrow8.x=80\)
\(\Rightarrow x=80:8\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
c) \(2\left|x-1\right|+5=15\)
\(\Rightarrow2\left|x-1\right|=15-5\)
\(\Rightarrow2\left|x-1\right|=10\)
\(\Rightarrow\left|x-1\right|=10:2\)
\(\Rightarrow\left|x-1\right|=5\)
Xét trường hợp 1: \(x-1=5\)
\(\Rightarrow x=5+1\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-1=-5\)
\(\Rightarrow x=\left(-5\right)+1\)
\(\Rightarrow x=-5+1\)
\(\Rightarrow x=-\left(5-1\right)\)
\(\Rightarrow x=-4\)
Vậy \(x=6\) hoặc \(x=-4\)
Bài 1:
\(a)\left(-175\right)-123+3+175+\left(-3\right)\)
\(=\left[\left(-175\right)+175\right]+\left[\left(-3+3\right)\right]-123\)
\(=0+0-123\)
\(=-123\)
\(b)17.85+15.17-150\)
\(=17.\left(85+15\right)-150\)
\(=17.100-150=1700-150=1550\)
\(c)176:\left(4.5^2-3.2^2\right)\)
\(=176:\left(4.25-3.4\right)\)
\(=176:\left(100-12\right)=176:88=2\)
Bài 2:
\(a)8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=20\)
\(b)20+2^3.x=5^2.4\)
\(\Rightarrow20+8.x=25.4\)
\(\Rightarrow20+8.x=100\)
\(\Rightarrow8.x=100-20\)
\(\Rightarrow8.x=80\)
\(\Rightarrow x=80:8=10\)
\(c)2\left|x-1\right|+5=15\)
\(\Rightarrow2\left|x-1\right|=15-5\)
\(\Rightarrow2\left|x-1\right|=10\)
\(\Rightarrow\left|x-1\right|=10:2\)
\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
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1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
\(3x-12=x-17\)
2x = -5
x = -5/2
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