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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 !
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
Câu hỏi của Lê Xuân Phú - Toán lớp 7 - Học toán với OnlineMath
vi \(\frac{-2009}{2010}>\frac{-2010}{1010}=1\)
\(\frac{2010}{-2009}=\frac{-2010}{2009}<\frac{-2009}{2009}=-1\)
=> x<y
violympic vòng 2