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\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\cdot9}=2\)
\(\dfrac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\dfrac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^2\cdot\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\dfrac{3^2\cdot\left(2^2\right)^2\cdot\left(2^{16}\right)^2}{11\cdot2^2\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{24}-2^{36}}\)
\(=\dfrac{3^2\cdot2^{34}}{11\cdot2^{24}-2^{36}}=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{11\cdot2^{24}-2^{12}\cdot2^{24}}\)
\(=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{\left(11-2^{12}\right)\cdot2^{24}}=\dfrac{3^2\cdot2^{10}}{11-2^{12}}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}=2\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{9.2^{36}}{2^{35}.\left(11-2\right)}=\frac{9.2^{36}}{2^{35}.9}=2\)
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 !!!
\(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\dfrac{\left(3.2^2.2^{16}\right)^2}{11.2^2.\left(2^2\right).11-\left(2^4\right)^9}\)
\(=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^2.2^{22}-2^{36}}\)
\(=\dfrac{3^2.2^4.2^{32}}{11.2^{24}-2^{36}}\)
\(=\dfrac{3^2.2^{34}}{11.2^{24}-2^{36}}\)
\(=\dfrac{3^2.2^{24}.2^{10}}{11.2^{24}-2^{12}.2^{24}}\)
\(=\dfrac{3^2.2^{24}.2^{10}}{\left(11-2^{12}\right).2^{24}}\)
\(=\dfrac{3^2.2^{10}}{11-2^{12}}\)
1.
\(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0\)
1 B=20182018.2017-20172017.2018
B=2018.10001.2017-2017.10001.2018
B=0
2 C=12+22+32+...+1002
C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)
C=1+2+1.2+3+2.3+...+100+99.100
C=(1+2+3+...+100)+(1.2+2.3+...+99.100)
C=[(1+100).100:2]+[(99.100.101):3]
C=5050+333300
C=338350