Question

Toán khó:

So sánh A và B, biết:

A= \(\dfrac{-21}{10^{2016}}+\dfrac{-12}{10^{2017}}\) và B= \(\dfrac{-12}{10^{2016}}+\dfrac{-21}{10^{2017}}\)

Answer 1

Ta có:A=\(\dfrac{-21}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\)

= \(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).

B=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-21}{10^{2017}}\)

=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2017}}\)+ \(\dfrac{-12}{10^{2017}}\)

Khi đó để so sánh A và B ta chỉ cần so sánh:\(\dfrac{-9}{10^{2016}}\)và \(\dfrac{-9}{10^{2017}}\)vì A và B cùng có:

\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).

Do:\(\dfrac{9}{10^{2016}}\)>\(\dfrac{9}{10^{2017}}\).

Suy ra:\(\dfrac{-9}{10^{2016}}\)<\(\dfrac{-9}{10^{2017}}\).

Từ đó ta suy ra được: A< B

Answer 2

TA CÓ:

A = \(-\dfrac{21}{10^{2016}}+-\dfrac{12}{10^{2017}}=\dfrac{-210+-12}{10^{2017}}=-\dfrac{222}{10^{2017}}\)

B = \(\dfrac{-12}{10^{2016}}+\dfrac{-21}{10^{2017}}=\dfrac{-120+-21}{10^{2017}}=\dfrac{-141}{10^{2017}}>\dfrac{-222}{10^{2017}}\)

=> B>A