Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.\(\frac{1001}{1000}>\frac{1000}{1000}=1=\frac{1003}{1003}>\frac{1002}{1003}\Rightarrow\frac{1001}{1000}>\frac{1002}{1003}\)
2.a) \(x=\frac{a-3}{2a}\left(a\ne0\right)\)
\(=\frac{1}{2}\left(1-\frac{3}{a}\right)\inℤ\)
\(\Leftrightarrow\hept{\begin{cases}1-\frac{3}{a}\inℤ\\1-\frac{3}{a}⋮2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3}{a}\inℤ\\\frac{3}{a}\equiv1\left(mod2\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\\\frac{3}{a}\equiv1\left(mod2\right)\end{cases}}\)
Ta có bảng :
| \(a\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
| \(\frac{3}{a}\) | \(3\) | \(-3\) | \(1\) | \(-1\) |
| \(1-\frac{3}{a}\) | \(-2\) | \(4\) | \(0\) | \(2\) |
| \(x\) | \(-1\) | \(2\) | \(0\) | \(1\) |
Vậy \(a\in\left\{\pm1;\pm3\right\}\)
b)Ta có:\(\frac{a+2009}{a-2009}=1+\frac{4018}{a-2009}\left(a\ne2009\right)\)
\(\frac{b+2010}{b-2010}=1+\frac{4020}{b-2010}\left(b\ne2010\right)\)
\(\Rightarrow\frac{4018}{a-2009}=\frac{4020}{b-2010}\)
\(\Rightarrow\frac{a-2009}{4018}=\frac{b-2010}{4020}\)
\(\Rightarrow\frac{a-2009}{2009}=\frac{b-2010}{2010}\)
\(\Rightarrow\frac{a}{2009}-1=\frac{b}{2010}-1\)
\(\Rightarrow\frac{a}{2009}=\frac{b}{2010}\)
Đặt A = \(\frac{2009^{2009}+1}{2009^{2010}+1}\)
B = \(\frac{2009^{2010}-2}{2009^{2011}-2}\)
Do 20092010- 2 < 20092011- 2 => \(B<1\)
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}<\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(1+2009^{2009}\right)}{2009\left(1+2009^{2010}\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=A\Rightarrow\)B < A
ta có:
B=(2009^2010-2)/(2009^2011-2)<1
=>(2009^2010-2)/(2009^2011-2)<(2009^2010-2)+2011/(2009^2011-2)+2011=(2009^2010+2009)/(2009^2011+2009)
=[2009*(2009^2009+1)]/[2009*(2009^2010+1)]=(2009^2009+1)/(2009^2010+1)=A
Vậy A=B
Đúng thì !
\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)
\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)
\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
vi \(\frac{-2009}{2010}>\frac{-2010}{1010}=1\)
\(\frac{2010}{-2009}=\frac{-2010}{2009}<\frac{-2009}{2009}=-1\)
=> x<y
violympic vòng 2