Tìm x biết \(\chi\cdot\left(\chi-2013\right)+\chi-2013=0\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B> \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\)
\(\Leftrightarrow\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(\left(x-\sqrt{x^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+2013}=-x+\sqrt{x^2+2013}\)
Chứng minh tương tự: \(x+\sqrt{x^2+2013}=-y+\sqrt{y^2+2013}\)
cộng vế theo vế ta được: \(x+y=-x-y\)
\(\Leftrightarrow x+y=0\Leftrightarrow x=-y\Leftrightarrow x^{2013}=-y^{2013}\)
\(\Leftrightarrow x^{2013}+y^{2013}=0\)
a,Ta có x =...
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3}+1}\right)\left(\sqrt{\sqrt{3}-1}\right)}\)
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)
x = \(\frac{\sqrt{3}.2}{\sqrt{3}}\)
x = 2
sau đó thay x=2 vào A nhé.
A=2014 !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2013}{4030}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
=>x+1=2015
=>x=2014
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 2000 < 5n < 2013
=> 5n = 2001 ; 2002
Vì các số trên không chia hết cho 5
Nên n không là số tự nhiên (vô lý)
Vậy không có số tụ nhiên n thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\\ \Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\\ \Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\left(1\right)\)
Tương tự: \(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\left(2\right)\)
Do đó: 2x=-2y
Suy ra: x=-y
Do đó:
\(x^{2013}+y^{2013}=\left(-y\right)^{2013}+y^{2013}=0\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 1 x 2 x 3 x ..... x 2012 x 2013 - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2013) x (2 x 4 x 6 x ..... x 2012) - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2011 x 2013) x (2 x 4 x 6 x ..... x 2012 - 1)
\(\left(x-2013\right)+x-2013=0\)
\(\left(x-2013\right)+\left(x-2013\right)=0\)
\(2\left(x-2013\right)=0\)
khi \(x-2013=0\Rightarrow x=2013\)
Vậy x=2013