Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)=\left(100+121+144\right):\left(169+196\right)=1\)
\(9!-8!-7!\cdot8^2=8!\left(9-1\right)-7!\cdot8^2=7!\cdot8^2-7!\cdot8^2=0\)
\(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}=\frac{9\cdot2^{36}}{2^{35}\cdot\left(11-2\right)}=\frac{9\cdot2^{36}}{2^{35}\cdot9}=2\)
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)
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 !!!
\(b)\)\(9!-8!-7!.8^2\)
\(=\)\(8!\left(9-1\right)-7!.8^2\)
\(=\)\(7!.8.8-7!.8^2\)
\(=\)\(7!.8^2-7!.8^2\)
\(=\)\(0\)
\(c)\)\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\)\(\frac{\left(2^2\right)^2.\left(2^{16}\right)^2.3^2}{2^{13}.\left(2^2\right)^{11}.11-\left(2^4\right)^9}\)
\(=\)\(\frac{2^4.2^{32}.3^2}{2^{13}.2^{22}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}\left(11-2\right)}\)
\(=\)\(\frac{2.3^2}{9}\)
\(=\)\(\frac{2.3^2}{3^2}\)
\(=\)\(2\)