Question

[Nguyen Le Quynh Trang]

Bài 1. So sánh A và B biết rằng :

a)  A = \(\frac{10^{15}+1}{10^{16}+1}\);  B = \(\frac{10^{16}+1}{10^{17}+1}\)

b)  A = \(\frac{3}{8^3}\)+\(\frac{7}{8^4}\);  B = \(\frac{7}{8^3}\)+\(\frac{3}{8^4}\)

c)  A = \(\frac{1}{11}\)+\(\frac{1}{12}\)+\(\frac{1}{13}\)+ ... +\(\frac{1}{19}\)+\(\frac{1}{20}\);  B = \(\frac{1}{2}\)

Answer 1

[Dương No Pro]

\(a.\)

\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)

\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)

\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)

\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)

\(10A=1+\frac{9}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)

\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)

\(10B=1+\frac{9}{10^{17}+1}\)

\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)

Answer 2

[Dương No Pro]

xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)

b 

\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)

Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B