Tìm x
( 2mũ 2 + 1 ) × ( x + 14 ) = 5 mũ 2 × 4 ( 2 mũ 5 + 2 mũ 2 + 2 ) : 2
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599 - 42 x 597 - 32 x 59
= 597.(52 - 42) - 32.59
= 597.(25 - 16) - 32.59
= 597.9 - 9.59
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
\(A=2^0+2^1+2^2+2^3+2^4+2^5+\dots+2^{100}\\=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+\dots+(2^{99}+2^{100})+2^0\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+\dots+2^{99}\cdot(1+2)+1\\=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{99}\cdot3+1\\=3\cdot(2+2^3+2^5+\dots+2^{99})+1\)
Vì \(3\cdot(2+2^3+2^5+\dots+2^{99})\vdots3\)
\(\Rightarrow 3\cdot(2+2^3+2^5+\dots+2^{99})+1\) chia \(3\) dư 1
hay số dư của phép chia \(A\) cho \(3\) là \(1\).
A=2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + (2^1 + 2^2) + (2^3 + 2^4) + ....+(2^99 + 2^100)
A=1 + 2.(1+2) + 2^3.(1+2)+....+2^99.(1+2)
A=1 + 2 . 3 + 2^3 . 3 +....+2^99 . 3
A=1 +3 .(2+2^3+..+2^99)
=> A:3 dư 1
12 + ( 5 + x ) = 20 5.22 + ( x + 3 ) = 52 23 + ( x + 3 ) = 52 43 - ( x - 2 ) = 52
17 + x = 20 5.4 + x + 3 = 25 8 + x + 3 = 25 64 - x + 2 = 25
x = 20 - 17 20 + 3 + x = 25 11 + x = 25 66 - x = 25
x = 3 23 + x = 25 x = 25 - 11 x = 66 - 25
x = 25 - 23 x = 14 x = 41
x = 2
Đăng nhìu v bn :) Đáng quan ngại đây :)
`@` `\text {Ans}`
`\downarrow`
`(2^2+1) \times (x+14) = 5^2 \times 4 + (2^5 + 3^2 + 7^2) \div 2`
` \Rightarrow (4+1) \times (x+14) = 5^2\times 2^2 + ( 32 + 9 + 49) \div 2`
`\Rightarrow 5 \times (x+14) = (5*2)^2 + (32+58) \div 2`
`\Rightarrow 5 \times (x+14) = 10^2+90 \div 2`
`\Rightarrow 5 \times (x+14) = 100 + 45`
`\Rightarrow 5 \times (x+14) = 145`
`\Rightarrow x+14 = 145 \div 5`
`\Rightarrow x+14=29`
`\Rightarrow x=29-14`
`\Rightarrow x=15`
Vậy, `x=15.`
\(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
\(\Rightarrow\left(4+1\right)\cdot\left(x+14\right)=25\cdot4+\left(32+9+49\right):2\)
\(\Rightarrow5\cdot\left(x+14\right)=100+45\)
\(\Rightarrow5x+70=145\)
\(\Rightarrow5x=75\)
\(\Rightarrow x=\dfrac{75}{5}=15\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
\(\dfrac{2^{13}+2^5}{2^{10}+2^5}\)
\(=\dfrac{2^5\cdot\left(2^8+1\right)}{2^5\cdot\left(2^5+1\right)}\)
\(=\dfrac{2^8+1}{2^5+1}\)