Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
Ta có:\(1-\frac{2010}{2010}=1-1=0\)
Tích\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)....\left(1-\frac{2011}{2010}\right)\)có chứa thừa số \(1-\frac{2010}{2010}=0\)
Vậy tích\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-\frac{2011}{2010}\right)=0\)
\(b)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Vậy \(\frac{2009^{2009}+1}{2009^{2010}+1}>\frac{2009^{1010}-2}{2009^{2011}-2}\)
Chúc bạn học tốt ~
Àk mình còn thiếu một điều kiện nữa xin lỗi nhé :
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Bạn thêm vào nhé
so sánh : cho A\(\frac{2010^{2011}+1}{2010^{2012}+1}\)
cho B =\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
Ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\)
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}\)
\(2010A=1+\frac{2009}{2010^{2012}+1}\)
Lại có:
\(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}\)
\(2010B=1+\frac{2009}{2010^{2011}+1}\)
Vì \(1+\frac{2009}{2010^{2012}+1}< 1+\frac{2009}{2010^{2011}+1}\)
nên 2010A < 2010B
hay A < B
Vậy A < B
Ta có: \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2011}+\frac{2012}{2010}}\)
\(=\frac{1}{2010\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)}+\frac{1}{2011\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}\right)}+\frac{1}{2012\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)}\)
\(=\frac{\frac{1}{2010}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}+\frac{\frac{1}{2011}}{\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}}+\frac{\frac{1}{2012}}{\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}=1\)
Mà \(\frac{2016}{2017}< 1\)
Vậy \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2010}+\frac{2012}{2011}}>\frac{2016}{2017}\)
dấu cần điền là : >
Vì kết quả của phép tính vế thứ 1 là 1
và phân số 2016/2017 bé hơn 1 nên ta điền dấu lớn
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
A=(1-1/2010).(1-1/2010).....(1-2011/2010)
A=1*(1/2010-2/2010-3/2010-...-2011/2010)
A=1/2010-2/2010-3/2010-...-2011/2010
rồi bạn bấm tiếp theo nha