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Ta có A-1=2016/20^10-1
B-1= 2016/20^10-3
Suy ra a-1<B-1=>A<B
Ta có:
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
Vì \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Ta có \(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(\Leftrightarrow A=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
\(\Leftrightarrow B=1+\dfrac{2}{20^{10}-3}\)
Vì 1=1 mà\(20^{10}-1>20^{10}-3\Rightarrow\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\Rightarrow1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
hay A < B
Vậy A < B
Ta có :
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{10^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)
Vì \(1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\Rightarrow A< B\)
Ta có:A=\(\dfrac{20^{10}+1}{20^{10}-1}\)>1\(\Leftrightarrow\)\(\dfrac{20^{10}+1}{20^{10}-1}< \dfrac{20^{10}+1-2}{20^{10}-1-2}\)=\(\dfrac{20^{10}-1}{20^{10}-3}\)=B
Vậy A<B
Ta có :
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
Vì \(1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
~ Chúc bn học tốt~
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\) (1)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\) (2)
vì \(20^{10}-1>20^{10}-3\)
nên \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\) (3)
từ (1), (2) và (3) suy ra A<B
Ta có :
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1\dfrac{2}{20^{10}-3}\)
Mà \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\Rightarrow A< B\)
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1\dfrac{2}{20^{10}-1}\) (đổi ra hỗn số)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1\dfrac{2}{20^{10}-3}\)
Do \(20^{10}-1>20^{10}-3\) nên \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\Rightarrow1\dfrac{2}{20^{10}-1}< 1\dfrac{2}{20^{10}-3}\Leftrightarrow A< B\)
Đáp số: A <B
Ta có:
\(A=\frac{20^{10}+1}{20^{10}-1}=1\)
\(B=\frac{20^{10}-1}{20^{10}-3}=1\)
Vậy A và B bằng nhau
Tính A và B rồi ta so sánh:
A = \(\frac{20^{10}+1}{20^{10}-1}\) = \(1\)
B = \(\frac{20^{10}-1}{20^{10}-3}\) = \(1\)
Mà \(1\) = \(1\)
Nên: A = B
\(A=\dfrac{20^{10}-1+2016}{20^{10}-1}=1+\dfrac{2016}{20^{10}-1}\)
\(B=\dfrac{20^{10}-3+2016}{20^{10}-3}=1+\dfrac{2016}{20^{10}-3}\)
mà \(20^{10}-1>20^{10}-3\)
nên A<B