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a) 2^x . 16^2 = 1024 b) 64 . 4^x = 16^8 c) 2^x = 16
=> 2^x . 256 = 1024 => 64 . 4^x = (4^2) ^ 8 => 2^x = 2^4
=> 2^x = 1024 : 256 => 4^3 . 4^x = 4^16 => x = 4
=> 2^x = 4 => 4^x = 4^16 : 4^3
=> 2^x = 2^2 => 4^x = 4^13
=> x = 13
=> x = 2
a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)
b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)
c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)
a) Ta có: 5 .(x+4) = 123 - 38
5 .(x+4) = 85
x+4 = 85 : 5
x+4 = 17
x = 17 - 4
x = 13
Câu b: Tương tự
a) 5 . ( x + 4 ) = 123 - 38
5 . ( x + 4 ) = 85
x + 4 = 85 : 5
x + 4 = 17
x = 17 - 4
x = 13
b ) \(3.x-16=2.7^4:7^3\)
3 . x - 16 = 2 . 7
3 . x - 16 = 14
3 . x =14 + 16
3 . x = 30
x = 30 : 3
x = 10
k mik nhé các bn I LVE YOU VERY MUCH
\(\frac{x+22}{11}+\frac{x+23}{12}=\frac{x+24}{13}+\frac{x+25}{14}\)
\(\Leftrightarrow\left(\frac{x+22}{11}+1\right)+\left(\frac{x+23}{12}+1\right)=\left(\frac{x+24}{13}+1\right)+\left(\frac{x+25}{14}+1\right)\)
\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}=\frac{x+37}{13}+\frac{x+39}{14}\)
\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}-\frac{x+37}{13}-\frac{x+39}{14}=0\)
\(\Leftrightarrow\frac{2184\cdot\left(x+33\right)+2002\cdot\left(x+35\right)-1848\cdot\left(x+37\right)-1716\cdot\left(x+39\right)}{24024}=0\)
\(\Leftrightarrow2184x+72072+2002x+70070-1848x-68376-1716x-66924=0\)
\(\Leftrightarrow622x+6842=0\)
\(\Leftrightarrow x=-11\)
a) A = 4 + 4 + 8 + 16 + ...... + 1048576
2A = 8 + 8 + 16 + ...... + 1048576 + 2.1048576
2A - A = (8 + 8 + 16 + ...... + 1048576 + 2.1048576) - (4 + 4 + 8 + 16 + ...... + 1048576)
A = 2.1048576 + 8 - 4 - 4
A = 2.1048576 = 2097152
b) (x + 1) + (x + 2) + ...... + (x + 100) = 5750
x + 1 + x + 2 + ...... + x + 100 = 5750
100x + (1 + 2 + 3 + ..... + 100) = 5750
Ta có :
1 + 2 + 3 + ..... + 100 = 5050
=> 100x + 5050 = 5750
=> 100x = 200
=> x = 2
Ta có :
x/5=9/3=>x=9.5/3=15
Mà :x+y=16
=>15+y=16
=> y=16-15
=> y=1
Vậy :x=15 và y=1
Vương Nguyên