\(10^6\) tận cùng là 0 \(=>10^6+2\) tận cùng là 2 \(=>10^6+2\) chia hết cho 2

sao ko dung f(x) ma viet

\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^9+2^{10}\)

a=\(\left(2+2^2\right)+2^2.\left(2+2^2\right)+..+2^8\left(2+2^2\right)\)

a=\(\left(2+2^2\right).\left(1+2^2+..+2^8\right)\)

a=\(6.\left(1+2^2+2^4+2^6+2^8\right)\)

chia het cho 3

2³+2⁴+2⁵+2⁶+2⁹⁸+2007⁰ đề thế hả bạn

gọi tổng đó là A:

A=1+2²+2³+2⁴+...+2⁹⁸

2A=2²+2³+2⁴+...+2⁹⁸+2⁹⁹

2A-A=2⁹⁹-1

A=2⁹⁹-1

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

Ta có:

$\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}$

$=\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+2^5\cdot5^7}$

$=\dfrac{2^{11}\cdot5^7+2^8\cdot5^{10}}{2^8\cdot5^4+2^5\cdot5^7}$

$=\dfrac{2^8\cdot5^7(2^3+5^3)}{2^5\cdot5^4(2^3+5^3)}$

$=\dfrac{2^3\cdot5^3}{1}$

$=8\cdot125$

$=1000.$

co ai giai dc ko cau 60 % .....

$\textbf{Ta có:}$

$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$

$=\dfrac{-25-45}{125+225}$

$=\dfrac{-70}{350}$

$=-\dfrac15.$

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.$

$B=\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}$

$=\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+2^5\cdot5^7}$

$=\dfrac{2^{11}\cdot5^7+2^8\cdot5^{10}}{2^8\cdot5^4+2^5\cdot5^7}$

$=\dfrac{2^8\cdot5^7(2^3+5^3)}{2^5\cdot5^4(2^3+5^3)}$

$=2^{8-5}\cdot5^{7-4}$

$=2^3\cdot5^3$

$=8\cdot125$

$=1000.$

(5^7 + 5^9) : (6^8+6^10)x (2^4-4^2)

=(5^7 + 5^9) : (6^8+6^10)x0

= 0

(5⁷+5⁹):(6⁸+6¹⁰)(2⁴-4²)=0