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Ta có: B = \(\frac{2012+2013}{2013+2014}=\frac{2012}{2013+2014}+\frac{2013}{2013+2014}\)
Mà : \(\frac{2013}{2014}>\frac{2013}{2013+2014}\)và \(\frac{2012}{2013}>\frac{2012}{2013+2014}\)
=> A > B
k nhé
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 \(\frac{2012^{2013}}{2013^{2013}}< \frac{2012^{2012}}{2013^{2012}}\)
\(\Rightarrow\frac{2012^{2013}}{2013^{2013}}+1< \frac{2012^{2012}}{2013^{2012}}+1\)
\(\Rightarrow\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}< \left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\)
a2014+b2014+c2014=1
a2015+b2015+c2015=1
=>a2014+b2014+c2014=a2015+b2015+c2015=1
=>a=b=1
=>A=3
ta có A+B
=\(\sqrt{2013}-\sqrt{2012}+\sqrt{2014}-\sqrt{2013}\) =\(-\sqrt{2012}+\sqrt{2014}\) (1)
vì (1)>0 nên A+B>0 hay A>B
A=\(\sqrt{2013}\)- \(\sqrt{2012}\) =\(\frac{1}{\sqrt{2013}+\sqrt{2012}}\)
B=\(\sqrt{2014}-\sqrt{2013}=\frac{1}{\sqrt{2014}+\sqrt{2013}}\)
sao sanh \(A=\frac{1}{\sqrt{2013}+\sqrt{2012}}>\frac{1}{\sqrt{2014}+\sqrt{2013}}\)
h cho minh nhieu nha
Ta có: \(A^2=4026+2\cdot\sqrt{2012\cdot2014}\)
\(B^2=4026+4026=4026+2\cdot\sqrt{2013^2}\)
mà \(2012\cdot2014< 2013^2\)
nên A<B
\(a^{2012}+b^{2012}+c^{2012}\ge3\sqrt[3]{\left(abc\right)^{2012}}=3\)
\(\Rightarrow\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\le\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge-\dfrac{1}{3}\)
Lại có:
\(a^{2013}+a^{2013}+...+a^{2013}\left(\text{2012 số hạng}\right)+1\ge2013\sqrt[2013]{\left(a^{2013}\right)^{2012}}=2013.a^{2012}\)
\(\Rightarrow2012.a^{2013}+1\ge2013.a^{2012}\)
Tương tự: \(2012.b^{2013}+1\ge2013.b^{2012}\) ; \(2012.c^{2013}+1\ge2013.c^{2012}\)
Cộng vế với vế:
\(\Rightarrow a^{2013}+b^{2013}+c^{2013}\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012}\)
\(\Rightarrow A\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012\left(a^{2012}+b^{2012}+c^{2012}\right)}=\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{3}=1\)
\(A_{min}=1\) khi \(a=b=c=1\)
Bạn tham khảo lời giải tại đây:
Câu hỏi của miumiucute - Toán lớp 9 | Học trực tuyến
\(A=\frac{2013^{2014}+1}{2013^{2015}+1}\)
\(\Rightarrow2013A=\frac{2013\left(2013^{2014}+1\right)}{2013^{2015}+1}=\frac{2013^{2015}+2013}{2013^{2015}+1}\)(1)
\(B=\frac{2013^{2012}+1}{2013^{2013}+1}\)
\(\Rightarrow2013B=\frac{2013\left(2013^{2012}+1\right)}{2013^{2013}+1}=\frac{2013^{2013}+2013}{2013^{2013}+1}\)(2)
Từ (1) và (2) => A<B
Hinh nhu minh hoc ca hai bang nhau ma