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Ta có : B = 202 - 192 + 182 - 172 + ..... + 22 - 12
=> B = (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + ..... + (2 - 1)(2 + 1)
=> B = 39 + 35 + 31 + ..... + 3
Số số hạng của dãy trên là :
(39 - 3) : 4 + 1 = 10 (số)
Tổng B là :
(39 + 3) x 10 : 2 = 210
Vậy B = 210
Ta có : \(C=\left(15^4-1\right)\left(15^4+1\right)-3^8.5^8\)
\(\Rightarrow C=\left(15^4\right)^2-1-15^8\)
\(\Rightarrow C=15^8-1-15^8\)
=> C = -1
Vậy C = - 1
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\(M=-1^2+2^2-3^2+4^2-...-17^2+18^2-19^2+20^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(18-17\right)\left(18+17\right)+\left(20-19\right)\left(20+19\right)\)
\(=3+7+...+35+39\)
\(=210\)
M = -12 + 22 - 32 + 42 -...- 172 + 182 - 192 + 202
= (2-1)(2+1) + (4-3)(4+3) +...+ (18-17)(18+17 + (20-19)(20+19)
= 3+7 +...+ 35 + 39
= 210
Chúc Bạn đạt thành tích cao trong năm học mới
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=(20^2-19^2)+(18^2-17^2)+.....+(4^2-3^2)+(2^2-1^2)
=(20+19)(20-19)+(18+17)(18-17)+.....+((4+3)(4-3)+(2+1)(2-1)
=39+35+.....+7+3
=(3+39)10/2=210
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Nhân hết ra rồi rút gọn thôi bạn:
\(A=\left(2^9+2^7+1\right)\left(2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}+2^9-2^7+1\right).\)
\(=2^{32}-2^{30}+2^{28}-2^{26}+2^{23}-2^{19}+2^{18}-2^{16}+2^9\)\(+2^{30}-2^{28}+2^{26}-2^{24}+2^{21}-2^{17}+2^{16}-2^{14}+2^7+2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}\)\(+2^9-2^7+1\)
\(=2^{32}+\left(2^{23}+2^{23}-2^{24}\right)+\left(2^{18}-2^{17}-2^{17}\right)+\left(2^9+2^9-2^{10}\right)+1=2^{32}+1\)
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mik ko chép lại đề bài nha
a) = (123)2- 12- (36. 46)
= (126-1)- (3.4)6
= 126-1-126
= -1
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ĐẶT
\(C=\text{ (20² + 18² + 16² + ... + 4² + 2²) - (19² + 17² + 15² + ... + 3² + 1²) }\)
\(C=\text{ 20² + 18² + 16² + ... + 4² + 2² - 19² - 17² - 15² - ... - 3² - 1² }\)
\(C=\text{ (20² - 19²) + (18² - 17²) + (16² - 15²) + .... + (4² - 3²) + (2² - 1²) }\)
\(C=\text{(20 + 19).(20 - 19) + (18 + 17).(18 - 17) + (16 + 15).(16 - 15) + .... + (2 + 1).(2 - 1) }\)
\(C=\text{ 20 + 19 + 18 + 17 + 16 + ..... + 2 + 1 }\)
\(C=\dfrac{20.\left(20+1\right)}{2}\)
\(C=210\)
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a) \(127^2+146.127+73^2=127^2+2.73.127+73^2=\left(127+73\right)^2=40000\)b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^8-1\right)=1\)
c) \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=100+99+98+97+...+2+1\)
\(=\dfrac{100\left(100+1\right)}{2}=5050\)
d) \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\) \(=20^2-19^2+18^2-17^2+16^2-15^2+...+4^2-3^2+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(2-1\right)\left(2+1\right)\)\(=20+19+18+17+...+2+1\)
\(=\dfrac{20\left(20+1\right)}{2}=210\)
e) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)
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a) Áp dụng hằng đẳng thức ta đc:
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=199+195+191+...+3\)
\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)
a) Áp dụng hằng đẳng thức ta đc:
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=199+195+191+...+3\)
\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)
b) mk nghĩ bước đầu tiên là phải bỏ ngoặc:
\(=20^2+18^2+16^2+...4^2+2^2-19^2-17^2-....-3^2-1^2\)
\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(4^2-3^2\right)-1^2\)
\(=\left(20+19\right)\left(20-19\right)+\left(18+17\right)\left(18-17\right)+...+\left(4-3\right)\left(4+3\right)-1\)
\(=\left(39+35+31+...+7\right)-1\)
\(=\left(\left[\left(39-7\right):4+1\right]\cdot\left(39+7\right):2\right)-1=207-1=206\)