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\(\frac{11.3^{32}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{39}-3^{30}}{4.3^{28}}=\frac{3^{30}.\left(11.3^9-1\right)}{4.3^{28}}=\frac{9.\left(11.3^9-1\right)}{4}=487152\)
\(a=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{\left(2^2.3^{28}\right)}=\frac{11.3^{29}-3.^{30}}{2^2.328}\)
\(=\frac{3^{28}\left(11.3-3^2\right)}{2^2.3^{28}}=\frac{33-9}{4}=6\)
Bài 35 :
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)
\(A=\frac{2^8.2^2.98}{2^8.104}\)
\(A=\frac{2^8.4.98}{2^8.4.26}\)
\(A=\frac{49}{13}\)
Vậy \(A=\frac{49}{13}\)
\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.\left(3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.3^{28}}\)
\(B=\frac{11.3^{29}-\left(3^2\right)^{15}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)
\(B=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)
\(B=\frac{3^{29}.8}{4.3^{28}}\)
\(B=\frac{3^{28}.3.4.2}{4.3^{28}}\)
\(B=3.2\)
\(B=6\)
Vậy B = 6
A = 2^10 . 13 + 2^10 . 65 / 2^8 . 104
= 2^10 ( 13 + 65 ) / 2^8 . 104 = 2^10 . 78 / 2^8 . 104 = 2^8 . 2^2 . 78 / 2^8 . 104 = 2^8 . 4 . 78 / 2^8 . 104 = 2^8 . 312 / 2^8 . 104
= 312/104
= 3
B = 11 . 3^22 . 3^7 - 9^15 / ( 2.3^14)^2
= 11 . 3^29 - (3^2)^15 / ( 3.2^14)^2
= 11 . 3^29 - 3^30 / ( 3. 2 )^28
= ( 8 + 3 ) . 3^29 - 3^30 / ( 3. 2)^28
= 8 . 3^29 + 3.3^29 - 3^30 / ( 3.2)^28
= 8 . 3^29 + 3^30 - 3^30 / ( 3 . 2)^28
= 8 . 3^29 / 3^28 . 2^28
= 2^3 . 3 / 2^28
= 3/ 2^25
ta co:A = 11.329-915/ (2.314)2
=11.329-(32)15/22.328
=11.329-330/22.328
=329.(11-3)/22.328
= 3.7/4
=21/4
\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(A=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(A=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)
\(A=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)
\(A=\frac{3^{29}.8}{4.3^{28}}=3.2=6\)
