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a)\(T=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
ta có \(2+1=2^2-1\)
\(T=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(T=\left(2^4-1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(T=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(T=2^{32}-1\)
bạn ơi nơi chổ mấy cái \(\left(2^2-1\right)\left(2^2+1\right)\)là nhân đa thức lại nha
b)
\(U=100^2-99^2+98^2-97^2+...+4^2-3^2+2^2-1^2\)
\(U=-1^2+2^2-3^2+4^2-...-97^2+98^2-99^2+100^2\)
\(U=2^2-1^2+4^2-3^2+...+98^2-97^2+100^2-99^2\)
\(U=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)(dùng hằng đẳng thức sô 3 nha)
\(U=3+7+...+199\)
\(U=1+2+3+\text{4+...+99+100}\)
số số hạng của U là :\(\left(100-1\right):1+1=100\) (số hạng)
tổng số số hạng của U là : \(\frac{\left(100+1\right).100}{2}=5050\)
à bạn coi lại cái đề nha đoạn sau hình như thiếu 2^2 thì phải
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
Bài 1:
a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)
b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(18^8-\left(18^8-1\right)=1\)
\(c,100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)
áp dụng công thức Gauss ta đc đáp án là:10100
d, mk khỏi ghi đề dài dòng:
\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:
\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)
\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)
\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)
1c,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)
A = -12 + 22 - 32 + 42 - ... - 992 + 1002
A = 1002 - 992 + ... + 42 - 32 + 22 - 12
A = (100 + 99).(100 - 99) + ... + (4 + 3).(4 - 3) + (2 + 1).(2 - 1)
A = 100 + 99 + ... + 4 + 3 + 2 + 1
\(A=\frac{\left(1+100\right).100}{2}=101.50=5050\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
2B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)...(332 + 1)
2B = (32 - 1)(32 + 1)(34 + 1)...(332 + 1)
2B = (34 - 1)(34 + 1)...(332 + 1)
2B = 364 - 1
\(B=\frac{3^{64}-1}{2}\)
A = 1002 - 992 + 982 - 972 + . . . + 22 - 12
= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)
= 199 + 195 + . . . + 3
= 5050
B = 3(22 + 1)(24 + 1) . . . (264 + 1) + 1
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1)(264 + 1) + 1
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (28 - 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (216 - 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (232 - 1)(232 + 1)(264 + 1) + 1
= (264 - 1)(264 + 1) + 1
= 2128 - 1 + 1
= 2128
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}...\frac{1-98^2}{98^2}.\frac{1-99^2}{99^2}\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{98^2-1}{98^2}.\frac{99^2-1}{99^2}\)
= \(\frac{\left(2-1\right).\left(2+1\right)}{2^2}.\frac{\left(3-1\right).\left(3+1\right)}{3^2}.\frac{\left(4-1\right).\left(4+1\right)}{4^2}...\frac{\left(98-1\right)\left(98+1\right)}{98^2}.\frac{\left(99-1\right)\left(99+1\right)}{99^2}\)
\(=\frac{\left(2-1\right).\left(3-1\right).\left(4-1\right)...\left(99-1\right)}{2.3.4...98.99}.\frac{\left(2+1\right).\left(3+1\right).\left(4+1\right)...\left(99+1\right)}{2.3.4...98.99}\)
\(=\frac{1.2.3....98}{2.3.4...98.99}.\frac{3.4.5...100}{2.3.4...98.99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{50}{99}\)
Chúc bạn học tốt !!!
M=(12+22+32)(22+32+42)......(982+992+1002)
e làm cho vuj thôi chứ ko có hứng để trình bày vs lại tính