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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) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)
ta có : \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+...+2^2-1^2\)
\(=\left(20^2-1^2\right)-\left(19^2-2^2\right)+\left(18^2-3^2\right)-...-\left(11^2-10^2\right)\)
\(=21.\left(20-1\right)-21\left(19-2\right)+21\left(18-3\right)-...-21\left(11-10\right)\)
\(=21.19-21.17+21.15-...-21.1\)
\(=21\left(19-17+15-13+...+3-1\right)\)
\(=21\left(2+2+...+2\right)=21.2.5=210\)
Ta có:\(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-15^2-...-3^2-1^2\)
\(=(20^2-19^2)+(18^2-17^2)+...+(4^2-3^2)+(2^2-1^2)\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(=20+19+18+17+...+4+3+2+1\)
\(=\dfrac{\left(20+1\right).20}{2}=\dfrac{21.20}{2}=210\)
mik ko chép lại đề bài nha
a) = (123)2- 12- (36. 46)
= (126-1)- (3.4)6
= 126-1-126
= -1
Bài 2: Bạn sử dụng các hằng đẳng thức đáng nhớ là ra.
a)
\(x^2+2x+1=(x+1)^2\)
b)
\(1-4x+4x^2=1^2-2.1.2x+(2x)^2=(1-2x)^2\)
c)
\(a^2+9-6a=a^2-2.3.a+3^2=(a-3)^2\)
\Leftrightarrow \left\{\begin{matrix}
\\
\end{matrix}\right.
=(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
ĐẶT
\(C=\text{ (20² + 18² + 16² + ... + 4² + 2²) - (19² + 17² + 15² + ... + 3² + 1²) }\)
\(C=\text{ 20² + 18² + 16² + ... + 4² + 2² - 19² - 17² - 15² - ... - 3² - 1² }\)
\(C=\text{ (20² - 19²) + (18² - 17²) + (16² - 15²) + .... + (4² - 3²) + (2² - 1²) }\)
\(C=\text{(20 + 19).(20 - 19) + (18 + 17).(18 - 17) + (16 + 15).(16 - 15) + .... + (2 + 1).(2 - 1) }\)
\(C=\text{ 20 + 19 + 18 + 17 + 16 + ..... + 2 + 1 }\)
\(C=\dfrac{20.\left(20+1\right)}{2}\)
\(C=210\)
Bài 11:
1) Sửa lại đề là: \(A=127^2+146.127+73^2\)
\(\Rightarrow A=127^2+2.127.73+73^2\)
\(\Rightarrow A=\left(127+73\right)^2\)
\(\Rightarrow A=200^2\)
\(\Rightarrow A=40000\)
Vậy \(A=40000.\)
2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)
\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(\Rightarrow B=18^8-\left(18^8-1\right)\)
\(\Rightarrow B=18^8-18^8+1\)
\(\Rightarrow B=0+1\)
\(\Rightarrow B=1\)
Vậy \(B=1.\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow D=\frac{3^{32}-1}{2}\)
\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)
\(=39+35+....+7+3\)
\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)