Câu 1:
a, $\frac{636363.37-373737.63}{1+2+3+...+2017}$ $=\frac{(63.1000+63.100+63).37-(37.10000+37.100+37).63}{1+2+3+...+2017}$ $=\frac{63.(10000+100+1).37-37.(10000+100+1).63}{1+2+3+...+2017}$ $=\frac{10101.(63.37-37.63)}{1+2+3+...+2017}=0$
b, *Cái này mình đang tính, nếu ra kết quả thì mình làm sau
Câu 3:
a, $A=3^1+...+3^{2006}$
$⇒3A=3^2+...+3^{2007}$
$⇒3A-A=-3+3^{2007}$
$⇒2A=3^{2007}-3$
$⇒A=$$\frac{3^{2007}-3}{2}$
b, Ta có: $2A+3=3^x$
$⇒3^{2007}-3+3=3^x$
$⇒3^{2007}=3^x$
$⇒x=2007$
Câu 4:
$A=\frac{2016^{2016}+1}{2016^{2017}+1}$=>$2016A=\frac{2016^{2017}+1+2015}{2016^{2016}+1}$ ⇒$2016A=1+\frac{2015}{2016^{2017}+1}$ (1)
$B=\frac{2016^{2015}+1}{2016^{2016}+1}$ ⇒$2016B=\frac{2016^{2016}+1+2015}{2016^{2016}+1}$ ⇒$2016B=1+\frac{2015}{2016^{2016}+1}$(2)
Từ (1) và (2)⇒$A<B$
