So sánh A và B,biết:
\(A=\left(1+\frac{1}{2013}\right)\left(1+\frac{1}{2013^2}\right).....\left(1+\frac{1}{2013^n}\right)\) (với n là số nguyên dương)
\(B=\frac{2013^2-1}{2012^2-1}\)
cho so A=\(\frac{2013+\frac{1}{2}}{\left(2012+\frac{1}{2}\right)^2+2013+\frac{1}{2}}\)
B=\(\frac{2013+\frac{1}{3}}{\left(2012+\frac{1}{3}\right)^2+2013+\frac{1}{3}}\)
so sanh A va B
So sánh \(\left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\) và \(\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}\)
Ta có \(\frac{2012^{2013}}{2013^{2013}}=\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}\)
Vì \(\frac{2012}{2013}< 1\)nên\(\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}< \frac{2012^{2012}}{2013^{2012}}.1=\frac{2012^{2012}}{2013^{2012}}\)
hay \(\frac{2012^{2013}}{2013^{2013}}< \frac{2012^{2012}}{2013^{2012}}\)
\(\Rightarrow\frac{2012^{2013}}{2013^{2013}}+1< \frac{2012^{2012}}{2013^{2012}}+1\)
\(\Rightarrow\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}< \left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\)
Cho A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)..\left(\frac{1}{2014^2}-1\right)\&B=\frac{1}{2}\) so sánh A và B
Ta có
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)
=> A<-1/2
Cho a,b,c là 3 số thực khác 0 và thỏa mãn:
\(\left\{{}\begin{matrix}a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)+2abc=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
Hãy tính giá trị của biểu thức: Q= \(\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}\)
gt \(\Rightarrow\left\{{}\begin{matrix}b\left(a^2+2ac+c^2\right)+ac\left(a+c\right)+b^2\left(a+c\right)=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\\a^{2013}+b^{2013}+c^{2013}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=0\Rightarrow a^{2013}+b^{2013}=0\\b+c=0\Rightarrow b^{2013}+c^{2013}=0\\a+c=0\Rightarrow a^{2013}+c^{2013}=0\end{matrix}\right.\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)
\(\Rightarrow Q=1\)
Cho ba số thực a,b,c thỏa mãn: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\). Tính giá trị của biểu thức:
\(P=\frac{17}{25}+\left(a^{2013}+b^{2013}\right)\left(b^{2013}+c^{2013}\right)\left(c^{2013}+a^{2013}\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a^{2013}+b^{2013}\right)\left(b^{2013}+c^{2013}\right)\left(c^{2013}+a^{2013}\right)=0\)
\(\Rightarrow P=\frac{17}{25}\)
\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Bài 1: Tính nhanh
\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{2011}\right)\)
Bài 2:
So sánh\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}\)với 3
\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)
\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)
\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)
\(=\frac{6}{2011}\)
Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) và B=\(\frac{-1}{2}\). Hãy so sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
Và B =\(\frac{1}{2}\)
So sánh A và B
A có: \(\frac{2014-2}{3-2}+1=2013\) ( thừa số )
Ta thấy mỗi thừa số của A đều có dạng \(\frac{1}{n^2}-1\)với \(n\inℕ^∗\)và \(n>1\)
Có \(\frac{1}{n^2}< 1\Rightarrow\frac{1}{n^2}-1< 1-1=0\)
=> Mỗi thừa số của A đều nhỏ hơn 0
=> A là tích của 2013 thừa số nhỏ hơn 0
Mà 2013 là số lẻ
=> A < 0
Mà B = \(\frac{1}{2}\)> 0
=> A < B