\(63^2-47^2\)

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

= \(16.110\)

= \(1760\)

a) \(63^2-47^2=\left(63-47\right)\left(63+47\right)=16\cdot110=1760\)

b) \(215^2-105^2=\left(215-105\right)\left(215+105\right)=35200\)

c) \(437^2-363^2=59200\)

d) \(537^2-463^2=74000\)

a,1/20

b,4/5

mình chưa học lớp 8

có cần bạn bình luận vậy ko 

\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63+47\right)\left(63-47\right)}{\left(215+105\right)\left(215-105\right)}=\frac{110\cdot16}{320\cdot110}=\frac{1}{20}\)

\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(473-363\right)\left(473+363\right)}{\left(573-463\right)\left(573+463\right)}=\frac{110\cdot836}{110\cdot1036}=\frac{836}{1036}=\frac{4\cdot209}{4\cdot234}=\frac{209}{234}\)

Trả lời:

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

\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437-363\right).\left(437+363\right)}{\left(537-463\right).\left(537+463\right)}=\frac{74.800}{74.1000}=\frac{4}{5}\)

Học tốt 

Áp dụng HĐT số 3 bạn nhé '-'

\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63+47\right)\left(63-47\right)}{\left(215+105\right)\left(215-105\right)}=\frac{110\cdot16}{320\cdot110}=\frac{16}{320}=\frac{1}{20}\)

\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437+363\right)\left(437-363\right)}{\left(537+463\right)\left(537-463\right)}=\frac{800\cdot74}{1000\cdot74}=\frac{800}{1000}=\frac{8}{10}=\frac{4}{5}\)

Ta có:63^2-47^2=3969-2209=1760

215^2-105^2=46225-11025=35200

ks nhé!Học tốt!

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

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

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

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

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

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

\(=\frac{16\cdot110}{110\cdot320}\)

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

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

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

các câu kia làm tương tự nha

Thanks bạn nhiều nhiều nha!

nhưng câu d với câu e mk khô hiểu lắm!

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

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.

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

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

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