1, tính :
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)..........\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
2, tìm x \(\frac{4x}{2x-\frac{1}{5}}>0\)
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4 tháng 3 2017
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
\(=\frac{-1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)......\left(-\frac{2003}{2004}\right)\left(-\frac{2004}{2005}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}.\frac{2004}{2005}\)
\(=\frac{1}{2005}\)
4 tháng 3 2017
Ta có : \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right).......\left(\frac{1}{2005}-1\right)\)
\(=-\frac{1}{2}.\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)........\left(-\frac{2004}{2005}\right)\)
\(=\frac{-1}{2}.\frac{2}{-3}.\frac{-3}{4}..........\frac{2004}{-2005}\)
\(=\frac{-1}{-2005}=\frac{1}{2005}\)
28 tháng 2 2017
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2004}{2\cdot3\cdot4\cdot...\cdot2005}\)
\(=\frac{1}{2005}\)
28 tháng 2 2017
\(=\frac{-1}{2}.\frac{-2}{3}....\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{1}{2005}\)
MN
24 tháng 1 2020
a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)
\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)
\(\Leftrightarrow6x+6-2x-2=5x+5-1\)
\(\Leftrightarrow6x-2x-5x=5-1-6+2\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)
\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)
\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)
\(\Leftrightarrow2x+6x=-7-5\)
\(\Leftrightarrow8x=-12\)
\(\Leftrightarrow x=-\frac{3}{2}\)
c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)
\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)
\(\Leftrightarrow\frac{x}{2}-\frac{2}{3}x+\frac{5}{6}x=-2+1-\frac{4}{3}+\frac{5}{3}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)
\(\Leftrightarrow x=-1\)
30 tháng 6 2017
a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times...\times\left(-\frac{2003}{2004}\right)\times\left(-\frac{2004}{2005}\right)\)
\(=\frac{1}{2005}\)
***
\(\frac{4x}{2x-\frac{1}{5}}>0\)
\(\Leftrightarrow\begin{cases}4x>0\\2x-\frac{1}{5}>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>0\\x>\frac{1}{10}\end{cases}\)
\(\Leftrightarrow x>\frac{1}{10}\)