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\(63^2-47^2\)
= \(\left(63-47\right)\left(63+47\right)\)
= \(16.110\)
= \(1760\)
\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63+47\right)\left(63-47\right)}{\left(215+105\right)\left(215-105\right)}=\frac{110\cdot16}{320\cdot110}=\frac{1}{20}\)
\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(473-363\right)\left(473+363\right)}{\left(573-463\right)\left(573+463\right)}=\frac{110\cdot836}{110\cdot1036}=\frac{836}{1036}=\frac{4\cdot209}{4\cdot234}=\frac{209}{234}\)
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
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 :
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Rightarrow4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
\(\Rightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Rightarrow4x=-2\)\(\Leftrightarrow x=-\frac{1}{2}\)
\(\frac{63^2-47^2}{215^2-105^2}=\) \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
\(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)
các câu kia làm tương tự nha
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.
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)
\(vì\:100< 104\:nên\:26^2-24^2< 27^2-25^2\\ hay\:A< B\)
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=\(\frac{4}{5}\)