\(63^2-47^2\)

= \(\left(63-47\right)\left(63+47\right)\)

= \(16.110\)

= \(1760\)

a,1/20

b,4/5

mình chưa học lớp 8

\(\frac{63^2-47^2}{215^2-105^2}=\)  \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

                           \(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)

các câu kia làm tương tự nha

Thanks bạn nhiều nhiều nha!

A=\(\frac{63^2-47^2}{215^2-105^2}\)

A=\(\frac{\left(63-47\right).\left(63+47\right)}{\left(215-105\right).\left(215+105\right)}\)

A=\(\frac{16.110}{110.320}\)

A=\(\frac{1760}{35200}\)

\(A=\frac{1}{20}\)

B=\(\frac{437^2-363^2}{537^2-463^2}\)

B=\(\frac{\left(437-363\right).\left(437+363\right)}{\left(537-463\right).\left(537+463\right)}\)

B=\(\frac{74.800}{74.1000}\)

B=\(\frac{4}{5}\)

A)16²/10²=8²/5²

B) 74²/69²

a) \(\dfrac{63^2-47^2}{215^2-105^2}\)

= \(\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

= \(\dfrac{16.110}{110.320}=\dfrac{16}{320}=\dfrac{1}{20}\)

b) \(\dfrac{437^2-363^2}{537^2-463^2}\)

= \(\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)

= \(\dfrac{74.800}{74.1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)

2)

A = \(26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)

B = \(27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)

Từ đó suy ra A < B

1.

\(a.\: \dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\\ =\dfrac{16.110}{110.320}=\dfrac{16}{320}=\dfrac{1}{20}\)

\(b.\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\\ =\dfrac{74.800}{74.1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)

2.

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)

\(vì\:100< 104\:nên\:26^2-24^2< 27^2-25^2\\ hay\:A< B\)

B3. 
a) =\(\frac{\left(63+47\right).\left(63-47\right)}{\left(215+105\right).\left(215-105\right)}\)               b) =\(\frac{\left(437+363\right).\left(437-363\right)}{\left(537+463\right).\left(537-463\right)}\)
  =\(\frac{110.16}{320.110}\)                                                   =\(\frac{800.74}{1000.74}\)
  =\(\frac{1}{20}\)                                                             =\(\frac{4}{5}\)

\(\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437+363\right)\left(437-363\right)}{\left(537+463\right)\left(537-463\right)}=\frac{800.74}{1000.74}=\frac{8}{10}=\frac{4}{5}\)