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\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\frac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)
\(=\frac{3^{29}.2^3}{2^2.3^{28}}\)
\(=3.2=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(a=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{\left(2^2.3^{28}\right)}=\frac{11.3^{29}-3.^{30}}{2^2.328}\)
\(=\frac{3^{28}\left(11.3-3^2\right)}{2^2.3^{28}}=\frac{33-9}{4}=6\)
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a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)
Vậy A = 3
b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)
\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)
Vậy B = 8
c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)
\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)
Vậy C = 6
d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)
Vậy D = 2
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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
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Đặt vế trái là A còn cái 6 . ( -14 )..... gì đó để nguyên ( gọi tạm là B đi)
Có 1 / 2 A = 2^3 / 7.15 + 2^3 / 15.23 +....+ 2^3 /55.63
Đến đây có khoảng cách hai thừa số dưới mẫu = tử nên ta có 1/2 A = 1/7 -1/15 +1/15 -1/23 + ... +1/55 -1/63
Khử hết cho nhau còn lại 1/7 - 1/63 = 8 / 63. Mặt khác đây là 1/2 A >> A = 4/63
Trở lại vế phải ta có B = 6. (-14) - 17. (-7) .(-2)/-22.28 = (-6).14-17.14/-22.28 ( đảo dấu 6 với -14 còn 1 cái 14 = -7. -2)
= 14 (17+ -6) /-22.28
= 14.11/-22.28
Rút 14 cho 28 và 11 cho -22 còn lại B = -1/4
Cả biểu thức = -1/4 + 4/63 = -47/252
\(A=\frac{11.2^{22}.2^7-4^{15}}{\left(3.2^{14}\right)^2}\)
\(A=\frac{11.2^{29}-2^{30}}{3^2.2^{28}}\)
\(A=\frac{2^{29}.\left(11-2\right)}{9.2^{28}}\)
\(A=\frac{2.9}{9}\)
\(A=2\)
vậy \(A=2\)