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\(=\frac{2^{10}\left(3^{31}+2^{30}.3^6\right)}{2^{11}\left(3^{31}+2^{30}.3^6\right)}=\frac{1}{2}\)
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.\left(2.3\right)^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^6.\left(3^{25}+2^{30}\right)}{2^{11}.3^6.\left(3^{25}+2^{30}\right)}=\frac{1}{2}\)
d) \(a.\left(-b\right).\left(-a\right)^2\left(-b\right)^3.\left(-a\right)^3.\left(-b\right)^4=-a^6b^8\)
Ta có 3^21>3^20
suy ra:3^20=(3^2)^10=9^10
2^31>2^30
suy ra:(2^3)^10=8^10
vì 8<9.Suy ra 2^31<3^21
\(A=\dfrac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}\)
\(A=\dfrac{2^{10}.3^6.3^{25}+2^{10}.2^{30}.3^6}{2^{11}.3^{25}.3^6+2^{11}.2^{30}.3^6}\)
\(A=\dfrac{2^{10}.3^6\left(3^{25}+2^{30}\right)}{2^{10}.3^6\left(3^{25}+2^{30}\right)}=1\)