So sánh :
\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10-1}}{20^{10}-3}\)
Những câu hỏi liên quan
so sánh\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
A=20^10+1/20^10-1=1*2/20^10-1
B=20^10-1/20^10+3=1*2/20^10-3
vi 20^10-1>20^10-3
Suy ra 2/20^10-1<2/20^10-3
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So sánh :
\(a.A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
\(b.A=\frac{10^{15}+1}{10^{16}+1}vàB=\frac{10^{16}+1}{10^{17}+1}\)
\(c.A=\frac{2013^5+2000}{2013^6+2000}vàB=\frac{2013^{10}+2000}{2013^{11}+2000}\)
Minh chi biet lam cau b thoi ak
b) Giai:
B=10^16+1 tren 10^17 +1 <10^16+1+9 tren 10^17+1+9
ma 10^16+1+9 tren 10^17+1+9 = 10^16+10 tren 10^17+10
=10(10^15+1) tren 10(10^16+1)
=10^15+1 tren 10^16+1 =A
=>A>B
Cho y kien voi!
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DÀI LẮM BN AK MK KO VIẾT NỔI
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mik cam on pn nhiu ,sau nay dug ai cmt lug tug nha....
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Rút gọn:
A =\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
So sánh :
\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
Y\(Sosanh:A=\frac{20^{10+1}}{20^{ }^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
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 = \(\frac{20^{10}+1}{20^{10}-1}và\frac{20^{10}-1}{20^{10}-3}\)
ta co:B=2010-1/2010-3>1
=>B>2010-1+2/2010-3+2=2010+1/2010-1=A
vay A<B
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so sánh \(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
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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