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:

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

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

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

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

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

$=8\cdot125$

$=1000.$

= 2031250 x 62145792 x 0

= 0

$(5^7+5^9)(6^8+6^{10})(2^4-4^2)$

$=5^7(1+5^2)\cdot6^8(1+6^2)\cdot(16-16)$

$=5^7\cdot26\cdot6^8\cdot37\cdot0$

$=0.$

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

\(a,\)\(\frac{2^5\times3^{12}\times7^8}{2^7\times3^{10}\times7^9}=\frac{3^2\times\left(2^5\times3^{10}\times7^8\right)}{2^2\times7\times\left(2^5\times3^{10}\times7^8\right)}\)\(=\frac{3^2}{2^2\times7}=\frac{9}{28}\)

\(b,\)Tương tự

$\textbf{a)}$

$\dfrac{2^5\cdot3^{12}\cdot7^8}{2^7\cdot3^{10}\cdot7^9}$

$=\dfrac{3^{12-10}}{2^{7-5}\cdot7^{9-8}}$

$=\dfrac{3^2}{2^2\cdot7}$

$=\dfrac{9}{28}.$

\(4^2\cdot120-4^3\cdot17+4^2\cdot34\)

\(=4^2\left(120-4\cdot17+34\right)\)

\(=16\left(120-68+34\right)\)

\(=16\cdot86=1088\)

\(7^3\cdot9+3^2\cdot7^4-45\cdot539\)

\(=7^3\cdot9+9\cdot7^4-45\cdot7^2\cdot11\)

\(=7^2\left(7\cdot9+9\cdot7^2-45\cdot11\right)\)

\(=49\left(63+9\cdot49-495\right)\)

\(=49\left(63+441-495\right)\)

\(=49\cdot9=441\)

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

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

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

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

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

$=2^3\cdot5^3$

$=8\cdot125$

$=1000.$

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.$

1: Ta có: \(\left(7x-3^2\right):5=8\)

\(\Leftrightarrow7x-9=40\)

hay x=7