a: \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)

\(=\frac{100+121+144}{169+196}=\frac{365}{365}=1\)

b: \(1\cdot2\cdot3\cdot\ldots\cdot9-1\cdot2\cdot3\cdot\ldots\cdot8-1\cdot2\cdot3\cdot\ldots\cdot7\cdot8^2\)

\(=1\cdot2\cdot3\cdot\ldots\cdot8\cdot\left(9-1-8\right)\)

=0

c: \(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^{18}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\frac{3^2\cdot2^{36}}{2^{35}\cdot\left(11-2\right)}=2\)

d: 1152-(374+1152)+(-65+374)

=1152-374-1152-65+374

=-65

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

a: \(\dfrac{10^2+11^2+12^2}{13^2+14^2}=\dfrac{100+121+144}{169+196}=\dfrac{365}{365}=1\)

c: \(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\cdot\left(11-2\right)}=2\)

b/1.2.3.....2013 - 1.2.3....2012-1.2.3...2012²

= (1.2.3...2012.2013-1.2.3...2012)-1.2.3...2012.2012

= 1.2.3...2012(2013-1-2012)

=1.2.3...2012.0=0

kkkkkkkkkkkkk giải giùm mình đi

(10*2+11*2+12*2):(13*2+14*2)=100+121+144):(169+196)=1

1.2.3.......9-1.2.3...8-1.2.3....8²

=1.2.3.......9-1.2.3.........8-1.2.3....8.8

= 1.2.3....8(9-1-8)

=1.2.3......8.0

=0

1234...7x(9-8-1)x8²

=0