Question

1.a) CMR 0<\(\frac{a}{b}\)<1;b>0;m>0

\(\frac{a}{m}\)< \(\frac{a+m}{b+m}\)

b) áp dụng để so sánh A= \(\frac{\text{2020^{2018}+1}}{2020^{2019}+1}\)và B=\(^{\frac{2020^{2017+1}}{2020^{2018+1}}}\)

Answer 1

Lời giải:
a) Sửa đề thành CM \(\frac{a}{b}< \frac{a+m}{b+m}\) mới đúng nhé.

\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{ab+am-b(a+m)}{b(b+m)}=\frac{m(a-b)}{b(m+b)}\)

Do $\frac{a}{b}< 1; b>0\Rightarrow a< b\Rightarrow a-b< 0$

$m>0; b>0\Rightarrow \frac{m}{b(m+b)}>0$

$\Rightarrow \frac{a}{b}-\frac{a+m}{b+m}=\frac{m(a-b)}{b(m+b)}< 0$

$\Rightarrow \frac{a}{b}< \frac{a+m}{b+m}$ (đpcm)

b)

\(0< B=\frac{2020^{2017+1}}{2020^{2018+1}}=\frac{2020^{2018}}{2020^{2019}}< 1\) do $0< 2020^{2018}< 2020^{2019}$

$1>0$

$\Rightarrow B< \frac{2020^{2018}+1}{2020^{2019}+1}$ (theo phần a)

Hay $B< A$