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So sánh\(A=\frac{2^{2006}+7}{2^{2004}+7}\)và\(B=\frac{2^{2003}+1}{2^{2001}+1}\)
A A > B
B A = B
C A < B
1.\(\frac{1}{a}=\frac{1}{6}+\frac{b}{3}=\frac{1}{6}+\frac{2b}{6}=2b+\frac{1}{6}=\frac{1}{a}\Rightarrow(2b+1)\cdot a=6=2b\cdot a+a=6=3a\cdot b=6\)
\(a\cdot b=\frac{6}{a}\)
\(3\cdot2\cdot b=6\Rightarrow a=2;b=1\)
2. \(\frac{a}{4}-\frac{1}{b}=\frac{3}{4}\)hay \(\frac{a}{4}-\frac{3}{4}=\frac{1}{b}=a-\frac{3}{4}=\frac{1}{b}=>(a-3)\cdot6=4\)
\(6a-18=4\)
\(6a=4+18=22\)
\(=>A\in\varnothing\)
Đúng nhé bạn
Ta có: - \(x\ge0;y\ge0\)
\(\Rightarrow\left|x+y\right|=\left|x\right|+\left|y\right|=x+y\)
- \(x\le0;y\le0\)
\(\Rightarrow\left|x+y\right|=\left|x\right|+\left|y\right|=-x-y=-\left(x+y\right)\)
- \(x\ge0;y\le0\)
\(\Rightarrow\left|x+y\right|=x+y< x< \left|x\right|+\left|y\right|\)
- \(x\le0;y\ge0\)
\(\Rightarrow\left|x+y\right|=x+y>x>\left|x\right|+\left|y\right|\)
\(\Leftrightarrowđpcm\)
\(2005a=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005b=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Ta thấy :\(2005^{2006}+1>2005^{2005}+1\)
\(\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\)
\(\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow2005a< 2005b\)
\(\Rightarrow a< b\)
Bài làm
Đặt a - b = x ; b - c = y ; c - a = z
=> x + y + z = 0
Ta có :
\(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{x+y+z}{xyz}\right)\)
=> \(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)( Vì x + y + z = 0 )
Vậy ta có đpcm