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\(B=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}=\frac{3^{29}.8}{4.3^{28}}=\frac{3.2.3^{28}.4}{4.3^{28}}=3.2=6\)
\(=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{14.2}}\)
\(=\frac{11.3^{29}-3.3^{29}}{2^2.3^{28}}\)
\(=\frac{\left(11-3\right).3^{29}}{2^2.3^{28}}\)
\(=\frac{8.3^{29}}{4.3^{28}}\)
\(=6\)
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
\(\frac{11.3^{2.2}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^4.3^7-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\frac{11.3^{11}-3^{30}}{2^2.3^{28}}\)
\(=\frac{3^{11}.\left(11-3^{19}\right)}{2^2.3^{28}}\)
\(=\frac{11-3^{19}}{2^2.3^{17}}\)
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Tham khảo đề và bài làm:
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.3^{14.2}}\)
\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3.8}{2^2}=6\)