So sánh: \(A=\frac{20^{10}+1}{20^{10}-1}\)với B\(\frac{20^{10}-1}{20^{10}-3}\)
Giúp mình nhé!
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So sánh A và B, biết :
A=\(\frac{20^{10}+1}{20^{10}-1}\) và B= \(\frac{20^{10}-1}{20^{10}-3}\)
Giúp mình nhá ^^ Mình tick cho, thanks nhiều :D
A = \(\frac{20^{10}+1}{20^{10}-1}=1\) B = \(\frac{20^{10}-1}{20^{10}-3}=1\)
Nên A = B
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\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
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Ta có : \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{\left(20^{10}-1\right)+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{\left(20^{10}-3\right)+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Do 2010 - 1 > 2010 - 3
\(\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
Hay A < B
Ai thấy tớ đúng k nha
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So sánh : \(A =\frac{20^{10} +1}{20^{10}-1} ; B =\frac{20^{10} -1}{20^{10} -3}\)
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> A < B
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so sánh \(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
Lời giải:
$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}=\frac{20^{10}-1}{20^{10}-3}=B$
Vậy $A< B$
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so sánh:
\(A=\frac{20^{10}+1}{20^{10}-1}\)với\(B=\frac{20^{10}-1}{20^{10}-3}\)
nhanh mik tik nha
Ta có \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
vÌ \(\frac{2}{20^{10}-3}>\frac{2}{20^{10}-1};1=1\Rightarrow1+\frac{2}{20^{10}-3}>1+\frac{2}{20^{10}-1}\Rightarrow A>B\)
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So sánh A và B:
\(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
So sánh A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
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
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So sánh : A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
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 \(A< B\)
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So sánh: \(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
Ta co B>1
=>A=20^10+1'20^10-1<20^10+1/2010-3+1
=>A<20^10+1/20^10-3+1
=>A<20^10-1/20^10-3
=>A<B
Vậy A<B
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So sánh :
\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10-1}}{20^{10}-3}\)
Ta có:
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Ta lại có:
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}\)
Hay A<B
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