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\(x^5+x^4+x^3+x^2+x+1=0\)
\(\Rightarrow x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^4+x^2+1\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
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1. 20182 - 2017.2019 = 20182 - ( 2018 - 1 )( 2018 + 1 ) = 20182 - ( 20182 - 12 ) = 20182 - 20182 + 12 = 1
2. 20012 = ( 2000 + 1 )2 = 20002 + 2.2000.1 + 12 = 4 000 000 + 4000 + 1 = 4 004 001
3. 19992 = ( 2000 - 1 )2 = 20002 - 2.2000.1 + 12 = 4 000 000 - 4000 + 1 = 3 996 001
4. 113 = ( 10 + 1 )3 = 103 + 3.102.1 + 3.10.12 + 13 = 1000 + 300 + 30 + 1 = 1331
5. 193 = ( 20 - 1 )3 = 203 - 3.202.1 + 3.20.1 - 13 = 8000 - 1200 + 60 - 1 = 6859
6. 57.63 = ( 60 - 3 )( 60 + 3 ) = 602 - 32 = 3600 - 9 = 3591
7. 953 + 3.952.5 + 3.95.52 + 53 = ( 95 + 5 )3 = 1003 = 1 000 000
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Ta có:
A=(100^2 -99^2)+(98^2 - 97^2)+(96^2 - 95^2)+.........+(2^2 - 1)
=(100-99)(100+99) + (98-97)(98+97) + (96-95)(96+95)+........+(2-1)(2+1)
=100+99+98+97+......+2+1=5050
Ở đây mình nhóm các hạng tử rồi AD hằng đẳng thức A^2 - B^2 = (A-B)(A+B)
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\(205^2-95^2=\left(205-95\right)\left(205+95\right)=110.300=33000\)
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\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\left(100+1\right).\frac{100-1}{2}=\frac{101.99}{2}=\frac{9999}{2}\)
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\(=\left(95^2-2\cdot60\cdot95+60^2\right)-30^2\\ =\left(95-60\right)^2-30^2=35^2-30^2=\left(35-30\right)\left(35+30\right)\\ =5\cdot65=325\)