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Công thức a/b >1 => a/b > a+n/b+n (a, b,n \(\in\) N*)
B = 2010-1/2010-3 > 1 nên B = 2010-1/2010-3 > 2010-1+2/2010-3+2 = 2010+1/ 2010-1 = A
Vậy A < B
C2:A=\(\frac{20^{10}+1}{20^{10}-1}\) B=\(\frac{20^{10}-1}{20^{10}-3}\)
A=\(\frac{20^{10}-1+2}{20^{10}-1}\) B=\(\frac{20^{10}-3+2}{20^{10}-3}\)
A=\(\frac{20^{10}-1}{20^{10}-1}\)+\(\frac{2}{20^{10}-1}\) B=\(\frac{20^{10}-3}{20^{10}-3}\)+\(\frac{2}{20^{10}-3}\)
A=1+\(\frac{2}{20^{10}-1}\) B=1+\(\frac{2}{20^{10}-3}\)
Vì 2010-1>2010-3=>\(\frac{2}{20^{10}-1}\)<\(\frac{2}{20^{10}-3}\)
=>1+\(\frac{2}{20^{10}-1}\)<1+\(\frac{2}{20^{10}-3}\)
=>A<B(đây là cách để đi thi còn cách kia làm cho nhanh thôi)
Vì B=\(\frac{20^{10}-1}{20^{10}-3}\)>1
Theo công thức \(\frac{a}{b}\)>1=>\(\frac{a+n}{b+n}\)>\(\frac{a}{b}\)
=>B>\(\frac{20^{10}-1+2}{20^{10}-3+2}\)
=>B>\(\frac{20^{10}+1}{20^{10}-1}\)=A
Vậy B>A
\(A=\frac{20^{10}+1}{20^{10}-1}>1\) và \(B=\frac{20^{10}-1}{20^{10}-3}>1\Rightarrow A=\frac{20^{10}+1}{20^{10}-1}>\frac{20^{10}+1-2}{20^{10}-1-2}=\frac{20^{10}-1}{20^{10}-3}=B\)
\(\Rightarrow A>B\)
ta có: 2010 + 1 > 2010 - 1
\(\Rightarrow A=\frac{20^{10}+1}{20^{10}-1}>1\)
Lại có: 2010 -1 < 2010 - 3
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}< 1\)
=> A > B
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1}{2^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3}{2^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
\(\text{Vì }2^{10}-1>2^{10}-3\text{ nên }1+\frac{2}{2^{10}-1}<1+\frac{2}{2^{10}-3}\)
Vậy A<B
Đặt \(x=20^{10}\)
Xét hiệu a - b: \(H=\frac{a+1}{a-1}-\frac{a-1}{a-3}=\frac{\left(a+1\right)\left(a-3\right)-\left(a-1\right)^2}{\left(a-1\right)\left(a-3\right)}=\frac{-2}{\left(a-1\right)\left(a-3\right)}< 0\)
Vậy a<b
Vì: A= 2010+1/ 2010 - 1 =1
B = 2010 -1/ 2010-3 = 1
Vì vaäy A= B
Mình nhầm, mình sẽ sủa lại:
Ta đã biết a/b >1 => a/b > a+n/b+n(a, b,n thuộc N*)
B = 2010-1/2010-3 > 1 nên B = 2010-1/2010-3 > 2010-1+2/2010-3+2
= 2010+1/ 2010-1 = A
Vậy B > A
A = 20^10 + 1 / 20^10 - 1 > 1
B = 20^10 - 1 / 20^10 - 3 > 1
Ta có: A = 20^10 + 1 / 20^10 - 1 > 20^10 + 1 - 2 / 20^10 - 1 - 2 = 20^10 - 1 / 20^10 - 3
Vậy A > B
Ta có: a/b > 1 => a/b > a+n/b+n { a,b,n thuộc N }
Vậy B= 20^10-1/20^10-3 > 20^10-1+2 / 20^10-3+2 = 20^10+1 / 20^10-1 = A
Kết luận B>A
Mình sẵn sàng trả lời mọi câu hỏi. Nhưng có những câu mình ko trả lời đc thì mong bạn thông cảm
mk có cách làm khác mấy bạn trên!!!!@@@
Các bạn toàn chép ở đằng sau sách đúng không!!
Ta có: A=\(\frac{20^{10}+1}{20^{10}-1}\)=\(1\frac{2}{20^{10}-1}\)(1)
B=\(\frac{20^{10}-1}{20^{10}-3}\)=\(1\frac{2}{20^{10}-3}\)(2)
Vì \(\frac{2}{20^{10}-1}\)<\(\frac{2}{20^{10}-3}\)(3)
Nên từ (1),(2),(3) suy ra A<B
Vậy A<B
MK làm đúng đó
Nguyễn Phương Thảo làm đúng rồi đó
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Ta có
A-1=(20^10+1/20^10-1)-1
=20^10+1/20^10-1-20^10-1/20^10-1
=20^10+1-(20^10-1)/20^10-1
=20^10+1-20^10+1/20^10-1
=2/20^10-1
Lại có:
B-1=(20^10-1/20^10-3)-1
=20^10-1/20^10-3-20^10-3/20^10-3
=20^10-1-(20^10-3)/20^10-3
=20^10-1-20^10+3/20^10-3
=2/20^10-3
Nhận thấy
2/20^10-1<2/20610-3(vì 20^10-1>20^10-3)
=>A-1<B-1
=>A<B
Ta có: \(B=\frac{20^{10}-1}{20^{10}-3}>1\)
\(\Rightarrow\)Ta có: \(B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1-2}{20^{10}-3-2}=\frac{20^{10}+1}{20^{10}-1}=A\)
\(\Rightarrow A< B\)
I don't know ok