\(63^2-47^2\)

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

= \(16.110\)

= \(1760\)

a,1/20

b,4/5

mình chưa học lớp 8

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

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

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 

A)16²/10²=8²/5²

B) 74²/69²

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

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

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

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.

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

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

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

Bài 1 :

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

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

Vì \(100< 104\Rightarrow A< B\)

Bài 2 :

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

\(\Rightarrow4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

\(\Rightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

\(\Rightarrow4x=-2\)\(\Leftrightarrow x=-\frac{1}{2}\)

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

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

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

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

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

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

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

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

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

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

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

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

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