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\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)
\(=\frac{2^{10}.78}{2^8.104}\)
\(=\frac{2^8.2^2.78}{2^8.104}\)
\(=\frac{2^8.4.78}{2^8.104}\)
\(=\frac{2^8.312}{2^8.104}\)
\(=\frac{2^8.3.104}{2^8.104}=3\)
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
a, \(A=2015.20162016-2016.20152015\)
\(A=2015.\left(2016.10001\right)-2016.20152015\)
\(A=\left(2015.10001\right).2016-20152015.2016\)
\(A=20152015.2016-20152015.2016\)
\(A=0\)
Vậy A = 0
b, \(B=\left(3.4.2^{16}\right)^2\div11.2^{13}.4^{11}-16^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.2^{22}-2^{36}\)
\(B=3^2.2^{36}\div11.2^{35}-2^{36}\)
\(B=3^2.2^{35}.2\div11.2^{35}-2.2^{35}\)
\(B=3^2.2\div9=9.2\div9=2\)
Vậy B = 2
c, \(C=2^{10}.13+2^{10}.65\div2^8.104\)
\(C=2^{10}.\left(13+65\right)\div2^8.104\)
\(C=2^{10}.78\div2^8.104\)
\(C=2^{10}.39\div2^8.13\)
\(C=39\div13=3\)
Vậy C = 3
Đề bài câu c sai mk sửa nhé là 28 ms tính đc k nó dư lắm !!!
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)
\(=\frac{2^{10}.78}{2^8.104}\)
\(=\frac{2^{10}.2.39}{2^8.2^3.13}\)
\(=\frac{2^{11}.39}{2^{11}.13}=3\)
a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)
b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)
c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=3\)
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=\frac{2^2.78}{104}=\frac{2^2.2.39}{2^3.13}=\frac{2^3.39}{2^3.13}=3\)