Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(Sửa \(cn-bm\rightarrow cn-dm\))
Ta có :
\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)
\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)
\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)
\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)
\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5\)
\(pt\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+3=0\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1+\dfrac{x+3}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{2004}{2004}+\dfrac{x+2}{2003}+\dfrac{2003}{2003}+\dfrac{x+3}{2002}+\dfrac{2002}{2002}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}+\dfrac{x+2005}{2002}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\right)=0\)
\(\Rightarrow x+2005=0\). Do \(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\ne0\)
\(\Rightarrow x=-2005\)
Ta có :
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.................+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+............+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+............+\dfrac{2}{2\left(n+1\right)}=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2003}{4008}\)
\(\dfrac{1}{n+1}=\dfrac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4007\) (Thỏa mãn \(n\in N\))
Vậy \(n=4007\) là giá trị cần tìm
~~Chúc bn học tốt~~
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right).\)
\(\Rightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(\Rightarrow A=\dfrac{1}{2004}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2004}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}...\dfrac{2003}{2004}\\ =\dfrac{1}{2004}\)
Cho A=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\); B=\(\dfrac{2003+2004}{2004+2005}\)
Ta có: B=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Vì: \(\dfrac{2003}{2004+2005}< \dfrac{2003}{2004}\)
\(\dfrac{2004}{2004+2005}< \dfrac{2004}{2005}\)
=>\(\dfrac{2003}{2004+2005}+\dfrac{2004}{2004+2004}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>\(\dfrac{2003+2004}{2004+2005}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>B<A
Vậy B<A
Ta thấy: \(\left|x\right|\ge0\forall x\)\(\Rightarrow\left|x\right|+2005\ge2005\forall x\)
\(\Rightarrow\dfrac{1}{\left|x\right|+2005}\le\dfrac{1}{2005}\forall x\)
\(\Rightarrow\dfrac{2004}{\left|x\right|+2005}\le\dfrac{2004}{2005}\forall x\Rightarrow A\le\dfrac{2004}{2005}\forall x\)
Đẳng thức xảy ra khi \(\left|x\right|=0\Leftrightarrow x=0\)
Vậy với \(x=0\) thì \(A_{Max}=\dfrac{2004}{2005}\)
Ta có :
\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2006}}\)
\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2005}}\)
Lại có :
\(-\dfrac{8}{10^{2006}}>\dfrac{-8}{10^{2005}}\Leftrightarrow M>N\)
\(B=\dfrac{2005}{x^m}+\dfrac{2003}{x^n}=\dfrac{2004}{x^m}+\dfrac{1}{x^m}+\dfrac{2004}{x^n}-\dfrac{1}{x^n}=A+\left(\dfrac{1}{x^m}-\dfrac{1}{x^n}\right)\)
\(\Rightarrow A< B\)
mình ko bt đúng hay sai nữa
\(A-B=\dfrac{1}{x^n}-\dfrac{1}{x^m}=\dfrac{x^m-x^n}{x^{m+n}}\)
+ Nếu m=n => A=B
+m>n => A>B
+m<n => A<B