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dùng hằng đẳng thức A^2 - B^2 = (A - B)(A + B) nhé phần b chuyển vế sang rồi dùng hđt là Okay
\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
b/ \(100^2+\left(100+3\right)^2+\left(100+5\right)^2+\left(100-6\right)^2\)
\(=100^2+100^2+100^2+100^2+4.100+9+25+36\)
\(=100^2+2.100+1+100^2-4.100+4+100^2-8.100+16+100^2+14.100+49\)
\(=\left(100+1\right)^2+\left(100-2\right)^2+\left(100-4\right)^2+\left(100+7\right)^2\)
Thực hiện phép chuyển vế, ta được:
(982-1002)+(1012-1032)=(942-962)+(1052-1072)
\(\Leftrightarrow\)(98-100)(98+100)+(101-103)(101+103)=(94-96)(94+96)+(105-107)(105+107)
\(\Leftrightarrow\)-2(98+100)-2(101+103)=-2(94+96)-2(105+107)
\(\Leftrightarrow\)98+100+101+103=94+96+105+107
\(\Leftrightarrow\)(100-96)+(98-94)=(107-103)+(105-101)
\(\Leftrightarrow\)4+4=4+4 (điều này hiển nhiên đúng)
Vậy 1002+1032+1052+942=1012+982+962+1072
\(VT-VP=\left(100^2-96^2\right)+\left(105^2-101^2\right)-\left(107^2-103^2\right)-\left(98^2-94^2\right)\)
\(=\left(100-96\right)\left(100+96\right)+\left(105-101\right)\left(105+101\right)-\left(107-103\right)\left(107+103\right)-\left(98-94\right)\left(98+94\right)\)
\(=4\left(196+206-210-192\right)=0\)
=> VT=VP
Xét hiệu , ta có :
1002 + 1032 + 1052 + 942 - ( 1012 + 982 + 962 + 1072 )
= 1002 + 1032 + 1052 + 942 - 1012 - 982 - 962 - 1072
= ( 1002 - 982 ) + ( 1032 - 1012 ) - ( 1072 - 1052 ) - ( 962 - 942 )
= ( 100 - 98 ).( 100 + 98 ) + ( 103 - 101 ).( 103 + 101 ) - ( 107 - 105 ). ( 107 + 105 ) - ( 96 - 94 ).( 96 + 94 )
= 2.198 + 2.204 - 2.212 - 2.190 = 2.( 198 + 204 - 212 - 190)
= 2.0 = 0
Vậy 1002 + 1032 + 1052 + 942 = 1012 + 982 + 962 + 1072.
Để \(100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2.\)
thì \(100^2+103^2+105^2+94^2-\left(101^2+98^2+96^2+107^2\right)\) phải bằng 0
\(\Rightarrow100^2+103^2+105^2+94^2-\left(101^2+98^2+96^2+107^2\right)=0\)
\(100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)
\(100^2-98^2+103^2-101^2+105^2-107^2+94^2-96^2\)
\(\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105^2-107^2\right)+\left(94^2-96^2\right)\)
\(\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)
áp dụng HĐT
\(\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(107-105\right)\left(107+105\right)-\left(96-94\right)\left(96+94\right)\)
\(=2\cdot198+2\cdot204-2\cdot212-2\cdot190\)
\(=2\left(198+204-212-190\right)\)
\(=2\cdot0=0\)(đpcm)
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.
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)
\(vì\:100< 104\:nên\:26^2-24^2< 27^2-25^2\\ hay\:A< B\)
\(a,\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}=\frac{16.110}{110.320}=\frac{1}{20}\)
\(b,\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}=\frac{74.800}{74.1000}=0,8\)
\(c,2^{32}-\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-2^{32}+1=1\)
\(d,100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)
\(=\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)
\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(107-105\right)\left(107+105\right)-\left(96-94\right)\left(96+94\right)\)
\(=2.198+2.204-2.212-2.190\)
\(=2\left(198+204-212-190\right)=2.0=0\)