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\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
\(b)\) 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
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)
\(A=\frac{2010}{1}+\frac{2009}{2}+...+\frac{2}{2009}+\frac{1}{2010}\)
\(A=1+\left(\frac{2009}{2}+1\right)+...+\left(\frac{2}{2009}+1\right)+\left(\frac{1}{2010}+1\right)\)
\(A=\frac{2011}{2011}+\frac{2011}{2}+...+\frac{2011}{2009}+\frac{2011}{2010}\)
\(A=\frac{2011}{2}+...+\frac{2011}{9}+\frac{2011}{10}+\frac{2011}{11}\)
\(A=2011.\left(\frac{1}{2}+...+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)\)
\(A=2011.B\)
Nên : \(\frac{A}{B}=\frac{2011.B}{B}=2011\)
Vậy \(\frac{A}{B}=2011\)
Tham khảo nha !!! Chúc bạn học tốt !!!
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:\(1-\frac{2010}{2010}=1-1=0\)
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\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-\frac{2011}{2010}\right)=0\)