\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
1,
\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)
\(|x- \frac{2}{7}|=\frac{1}{7}\)
<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)
Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)
Học tốt
\(5^{61}+25^{31}+125^{21}\)
\(=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)
\(=5^{61}\cdot5^{2\cdot31}\cdot5^{3\cdot21}\)
\(=5^{61}+5^{62}+5^{63}\)
\(=5^{61}\cdot\left(1+5+5^2\right)\)
\(=5^{61}\cdot\left(6+5^2\right)\)
\(=5^{61}\cdot\left(6+25\right)\)
\(=5^{61}\cdot31\)
Vì \(5^{61}\inℤ\)
\(\Rightarrow5^{61}\cdot31⋮31\)
\(\Rightarrow5^{61}+25^{31}+125^{21}⋮31\)
Vậy bài toán đã được chứng minh .
1)\(|x-\frac{2}{7}|=\frac{-1}{5}.\frac{-5}{7}\) 2)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2008}-1\right)\left(\frac{1}{2009}-1\right)\)
\(|x-\frac{2}{7}|=\frac{1}{7}\) \(=-\frac{1}{2}.-\frac{2}{3}.....-\frac{2007}{2008}.-\frac{2008}{2009}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{7}=\frac{1}{7}\\x-\frac{2}{7}=-\frac{1}{7}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{7}+\frac{2}{7}\\x=-\frac{1}{7}+\frac{2}{7}\end{cases}}}\) \(=\frac{1}{3}.\frac{3}{5}.....\frac{2005}{2007}.\frac{2007}{2009}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{7}\\x=\frac{1}{7}\end{cases}}\) \(=\frac{1}{2009}\)
1.\(\left|x-\frac{2}{7}\right|=\frac{-1}{5}\cdot\frac{-5}{7}\)
=> \(\left|x-\frac{2}{7}\right|=\frac{-1}{1}\cdot\frac{-1}{7}\)
=> \(\left|x-\frac{2}{7}\right|=\frac{1}{7}\)
=> x - 2/7 = 1/7 hoặc x - 2/7 = -1/7
=> x = 3/7 hoặc x = 1/7
2.\(\left[\frac{1}{2}-1\right]\left[\frac{1}{3}-1\right]...\left[\frac{1}{2008}-1\right]\left[\frac{1}{2009}-1\right]\)
\(=\left[-\frac{1}{2}\right]\left[-\frac{2}{3}\right]...\left[-\frac{2007}{2008}\right]\left[-\frac{2008}{2009}\right]\)
\(=\frac{(-1)(-2)...(-2008)}{2\cdot3\cdot2008\cdot2009}\)
\(=\frac{1\cdot2\cdot...\cdot2008}{2\cdot3\cdot...\cdot2008\cdot2009}=\frac{1}{2009}\)
3. Ta có : \(5^{61}+25^{31}+125^{21}\)
\(=5^{61}+\left[5^2\right]^{31}+\left[5^3\right]^{21}\)
\(=5^{61}+5^{62}+5^{63}\)
\(=5^{61}\left[1+5+5^2\right]\)
\(=5^{61}\cdot31⋮31\)
hay \(5^{61}+25^{31}+125^{21}⋮31(đpcm)\)
4.Ta đã biết với mọi x,y \(\inℚ\)thì \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Đẳng thức xảy ra khi xy >= 0
Ta có : \(A=\left|x-2011\right|+\left|x-200\right|=\left|x-2011\right|+\left|200-x\right|\ge\left|x-2011+200-x\right|=\left|-1811\right|=1811\)Vậy \(A\ge1811\), A đạt giá trị nhỏ nhất là 1811 khi \(200\le x\le2011\).
Cách khác :
Ta đã biết với mọi \(x\inℝ\)thì |x| = |-x| và \(\left|x\right|\ge x\)
Do đó suy ra \(A\ge1811\)
A đạt giá trị nhỏ nhất là 1811 khi \(2011-x\ge0\)và \(x-200\ge0\)tức là \(200\le x\le2011\)
