so sánh 10 mũ 1990 + phần 10 mũ 1991 +1 và 10 mũ 1991 +1 phần 10 mũ 1992 +1 06/10/2021 Bởi Katherine so sánh 10 mũ 1990 + phần 10 mũ 1991 +1 và 10 mũ 1991 +1 phần 10 mũ 1992 +1
Đặt $A=\frac{10^{1990}+1}{10^{1991}+1}$ $⇒10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{(10^{1991}+1)+9}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}$ $B=\frac{10^{1991}+1}{10^{1992}+1}$ $⇒10B=\frac{10^{1992}+10}{10^{1992}+1}=\frac{(10^{1992}+1)+9}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}$ Vì $10^{1991}+1<10^{1992}+1$ $⇒\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}$ $⇒1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}$ $⇒10A>10B$ $⇒A>B$. Bình luận
Đặt $A=\frac{10^{1990}+1}{10^{1991}+1}$
$⇒10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{(10^{1991}+1)+9}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}$
$B=\frac{10^{1991}+1}{10^{1992}+1}$
$⇒10B=\frac{10^{1992}+10}{10^{1992}+1}=\frac{(10^{1992}+1)+9}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}$
Vì $10^{1991}+1<10^{1992}+1$
$⇒\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}$
$⇒1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}$
$⇒10A>10B$
$⇒A>B$.