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A=1+2+3+4+...+2017+2018-2019
A=[(2018-1):1+1]-2019=2018
A= [(2018+1) x 2018 :2 =2037171
A= 2037171 - 2019
A=2035152
hok tốt
\(A=1+2+...+2017+2018-2019\)
\(A=\frac{\left(2018+1\right).2018}{2}-2019\)
\(A=2035152\)

nhanh nha mọi người. Ngày mai mình thi rồi. Ai giải nhanh nhất mình k cho!

Ta có: 22018-22016=22016(22-1)=2016\(\times\)3
vì 2016 \(\times\) 3 chia hết cho 3 nên 22018-22016 chia hết cho 3
ta có:
22018-22016=22016(22-1)=22016.3
Vì 22016.3 chia hết cho 3 nên 22018-22016


\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)
Tự so sánh
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)
Từ (1) và (2) suy ra 100A > 100B hay A > B
Phá ngoặc ra
(-2018)+[100-(-2018)-(-100)]
=(-2018)+100+2018+100
=200
\((-2018)+\left[100-(-2018)-(-100)\right]\)
\(=-2018+100+2018+100\)
\(=-2018+2018+100+100\)
\(=0+200=200\)