\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^6+2^{16}\cdot3^9}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^6\left(1+2^6\cdot3^3\right)}=\dfrac{3^2\cdot\left(-2\right)}{1+64\cdot27}=\dfrac{-18}{1729}\)

1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10

=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)

=9/10x(1/3+3/5+5/7+7/9)

9/10x(1/3+3/5)+(5/7+7/9)

=9/10x1/5+5/9

9/50+5/9

=10

Bn Long làm đúng rồi bn   nguyễn kim arica cứ làm theo cách đó là được .

Bn nào thấy đúng thì ủng hộ nha .

mk ko viết lại đề

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^4.7^4}{5^9.7^3+5^92^3.7^3}\)\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{7^3\left(5^{10}-5^4.7^4\right)}{7^3\left(5^9+5^9.2^3\right)}\)

\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5^{10}-5^4.7^4}{5^9+5^9.2^3}\)\(=\frac{3^4\left(3-1\right)}{3^4\left(3^2+3\right)}-\frac{5^4\left(5^6-7^4\right)}{5^4\left(5^5+5^5.2^3\right)}\)

\(=\frac{1}{6}-\frac{5^6-7^4}{5^5+5^5.2^3}\)cậu cứ phân tích từ từ

thanks bn nhìu nha !!

không vt lại đề

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.2^3.7^3}\)

\(=\frac{2^{12}\cdot3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+2^3\right)}\)

\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)

bạn chưa rút gọn 

(1+10) +(2+9) +....+ (5+6) + (10+1) +(9+2) +........+ (6+5)

= 11 x 10 = 110

Giải

 (1+9)+(2+8)...(6+4)+(7+3)...(9+1)+10+10

10*11=110