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ta thấy B>1 nên B=\(\frac{20^{10}-1}{20^{10}-3}\)>\(\frac{20^{10}-1+2}{20^{100}-3+2}\)=\(\frac{20^{10}+1}{20^{10}-1}\)=A
vậy B>A
nếu ko hiểu thì tham khảo trong SBT lớp 6 bài so sánh PS ấy
Vì \(20^{10}-1>20^{10}-3\)
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1+2}{20^{10}-3+2}=\frac{20^{10}+1}{20^{10}-1}=A\)
vậy \(A< B\)
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
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> A < B
Vì \(20^{10}-1>20^{10}-3\)
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1+2}{20^{10}-1+2}=\frac{20^{10}+1}{20^{10}-1}=A\)
\(\Rightarrow A< B\)
Ta có : \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{\left(20^{10}-1\right)+2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{\left(20^{10}-3\right)+2}{20^{10}-3}\)
\(A=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Do : \(20^{10}-1>20^{10}-3\)
\(\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
Vậy : \(A< B\)
Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
Vậy A < B