Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
$\textbf{Ta có:}$
$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$
$=\dfrac{-25-45}{125+225}$
$=\dfrac{-70}{350}$
$=-\dfrac15.$
Và
$B=\dfrac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}$
$=\dfrac{2^{12}3^{10}+2^{12}3^{10}\cdot5}{2^{11}3^{11}(2\cdot3-1)}$
$=\dfrac{2^{12}3^{10}(1+5)}{2^{11}3^{11}\cdot5}$
$=\dfrac{2^{12}3^{10}\cdot6}{2^{11}3^{11}\cdot5}$
$=\dfrac{2^2}{5}$
$=\dfrac45.$
=> $M=B-A$$=\dfrac45-\left(-\dfrac15\right)$
$=\dfrac55$$=1.$
Ta có:\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}\)
\(\Rightarrow A=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}\)
\(\Rightarrow A=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}\)
\(\Rightarrow B=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^8-1}>\frac{3}{10^8-3}\Rightarrow A>B\)
tìm số dư của
A= [22^6n+2(hai mũ hai mũ sáu n cộng hai) + 3]:7
B =[22^3n+1(hai mũ ba n cộng một)+3]:13
$\textbf{A)}$
$A=\left(2^{\,2^{6n+2}}+3\right):7.$
Ta có: $2^3\equiv1\pmod7.$
Lại có $6n+2\ge2\Rightarrow2^{6n+2}$ chia hết cho $4$, nên $2^{6n+2}\equiv1\pmod3.$
Suy ra $2^{2^{6n+2}}\equiv2^1\equiv2\pmod7.$
Vậy $2^{2^{6n+2}}+3\equiv2+3\equiv5\pmod7.$
$\Rightarrow$ Số dư là $5$.
$\textbf{B)}$
$B=\left(2^{\,2^{3n+1}}+3\right):13.$
Ta có: $2^{12}\equiv1\pmod{13}.$
Lại có $2^{3n+1}=2\cdot8^n.$
Vì $8^2\equiv12,\;8^3\equiv5,\;8^4\equiv1\pmod{12}$ nên $8^n\equiv4\pmod{12}$ khi $n\equiv2\pmod4$, do đó $2^{3n+1}\equiv8\pmod{12}.$
Suy ra $2^{2^{3n+1}}\equiv2^8\equiv256\equiv9\pmod{13}.$
Vậy $2^{2^{3n+1}}+3\equiv9+3\equiv12\pmod{13}.$
$\Rightarrow$ **Số dư là $12$.**
a, 20^2 - 6^ 2= 400 - 36 = 364
b,3^3 . 18 - 3^3 . 12 = 3^3 . ( 18-12) = 27 . 6 = 162
c, 39 . 213 + 87 . 39 = ( 213 + 87) . 39 = 300 . 39 = 110700
d, 80 - [ 130 - ( 12 - 4)^2] = 80 - [130 - (3^2) ] = 80 - 130 - 9 = -59
a, \(A=2^{2^{6n+2}}\)
Ta có: \(2^{6n+2}\equiv1\left(mod3\right)\)
\(\Rightarrow2^{6n+2}=3k+1\left(k\in Z\right)\)
\(\Rightarrow A=2^{3k+1}=4.2^{3k}=4.8^k\equiv4.1\equiv4\left(mod7\right)\)
Vậy A chia 7 dư 4
$\textbf{B)}$
$B=\left(2^{\,2^{3n+1}}+3\right):13.$
Ta có: $2^{12}\equiv1\pmod{13}.$
Lại có $2^{3n+1}=2\cdot8^n.$
Vì $8^2\equiv12,\;8^3\equiv5,\;8^4\equiv1\pmod{12}$ nên $8^n\equiv4\pmod{12}$ khi $n\equiv2\pmod4$, do đó $2^{3n+1}\equiv8\pmod{12}.$
Suy ra $2^{2^{3n+1}}\equiv2^8\equiv256\equiv9\pmod{13}.$
Vậy $2^{2^{3n+1}}+3\equiv9+3\equiv12\pmod{13}.$
$\Rightarrow$ **Số dư là $12$.**
\(a.2^6.\left(x-2\right)=104\)
\(x-2=104:2^6\)
\(x-2=1,652\)
\(x=1,625+2\)
\(x=3,625\)
\(b.2\times4^{x+1}=128\)
\(4^{x+1}=128:2\)
\(4^{x+1}=64\)
\(4^{x+1}=4^3\)
\(\Rightarrow x+1=3\)
\(x=3-1\)
\(\Leftrightarrow x=3\)
\(c.227-5\left(x+8\right)=3^6:3^3\)
\(227-5\left(x+8\right)=3^3\)
\(227-5\left(x+8\right)=27\)
\(5\left(x+8\right)=227-27\)
\(5\left(x+8\right)=200\)
\(x+8=200:5\)
\(x+8=40\)
\(x=40-8\)
\(x=32\)
ủng hộ mk nha, chắc đúng đó
cả tháng nay ms online lại
\(A=1+2^2+2^3+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(\Rightarrow A+1=2^{101}-1+1\)
\(\Rightarrow a+1=2^{101}\)
\(\Rightarrow2n+1=101\)
\(\Rightarrow2n=101-1\)
\(\Rightarrow2n=100\)
\(\Rightarrow n=100\div2\)
\(\Rightarrow n=50\)