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\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{6^{12}-6^{11}}=\frac{2^{12}.3^{10}.6}{6^{11}.\left(6-1\right)}=\frac{2^{12}.3^{10}.2.3}{6^{11}.\left(6-1\right)}=\frac{2^{13}.3^{11}}{6^{11}.5}=\frac{2^{11}.3^{11}.2^2}{6^{11}.5}=\frac{6^{11}.4}{6^{11}.5}=\frac{4}{5}\)
Bài2
a) ta có : 10^19 + 10^18 +10^17 = 10^17 (10^2+10+1)
= 10^17 . 111
Do 10 chia hết cho 5 nên 10^17 cũng chia hết cho 5. Mà 10^17 cũng chia hết cho 111
nên 10^17 chia hết cho 111x5 = 555 ( vì (111;5)=1)
Vậy 10^19 + 10^18 + 10^17 chia hết cho 555
b) Ta có : 7+7^2+7^3+7^4+...+7^84
= (7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^82+7^83+7^84)
= 7(1+7+7^2) + 7^4(1+7+7^2)+...+7^82(1+7+7^2)
= 7.57 + 7^4.57 +...+ 7^82.57
= 57(7.7^4....7^82) chia hết cho 57
Vậy 7+7^2+7^3+...+7^84 chia hết cho 57
ta co : A= ( 8^9+12/8^9+7) -1
= 5/8^9+7
B=(8^10+4/8^10-1)-1
=5/8^10-1
VI 8^9+7 < 8^10-1 NEN 5/8^9+7 > 5/8^10-1
VAY A > B
Ta có : A = ( 8^9+12/8^9+7) - 1
= 5/8^9 + 7
B = (8^10+4/8^10-1) - 1
= 5/8^10-1
VI 8^9 + 7 < 8^10 - 1 nên 5/8^9+7 > 5/8^10-1
\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{ \left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=2^8=256\)
Chúc học tốt :)
\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
\(M=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)
\(M=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(M=\frac{2^{20}.\left(2^{10}+1\right)}{2^{12}.\left(1+2^{10}\right)}\)
\(M=\frac{2^{20}}{2^{12}}\)
\(M=2^8=256\)
\(M=\frac{8^{10}+4^{10}=1.074.790.400}{8^4+4^{11}=4.198.400}\)
Vậy
\(\Rightarrow M=\frac{1.074.790.400}{4.198.400}\)
P/s; Ko chắc đâu nhé
8^10 + 4^10 / 8^4 + 4^11
= (2^3)^10 + (2^2)^10 / (2^3)^4 + (2^2)^11
= 2^30 + 2^20 / 2^12 + 2^22
= 2^20 . (2^10 + 1) / 2^12 . (1 + 2^10)
= 2^20 / 2^12
= 2^8
= 256
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\cdot\left(2^{10}+1\right)}{2^{12}\cdot\left(1+2^{10}\right)}=2^8=256\)