Question

[Kwon Ji-Yong]

Tính:

a) 63² - 47²/ 215² - 105²

b) 437² - 363²/ 537² - 463²

c) 2³² - (2 + 1). (2² + 1). (2⁴ + 1). (2⁸ + 1). (2¹⁶ + 1)

d) 100² + 103² + 105² + 94² - 101² - 98² - 96² - 107²

Answer 1

[lê thị hương giang]

\(a,\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}=\frac{16.110}{110.320}=\frac{1}{20}\)

\(b,\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}=\frac{74.800}{74.1000}=0,8\)

\(c,2^{32}-\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-2^{32}+1=1\)

\(d,100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)

\(=\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)

\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(107-105\right)\left(107+105\right)-\left(96-94\right)\left(96+94\right)\)

\(=2.198+2.204-2.212-2.190\)

\(=2\left(198+204-212-190\right)=2.0=0\)