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24 tháng 6 2015

1)\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)....\left(\frac{1}{2008}-1\right).\left(\frac{1}{2009}-1\right)=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2008}{2009}\right)=\frac{1.2.3...2008}{2.3.4....2009}=\frac{1}{2009}\)

2)\(A=\frac{x-7}{2}\)

Do 2>0 =>A>0 <=>x-7>0<=>x>7

Vậy x>7 thì A>0

3)\(A=\frac{x+3}{x-5}\)

Do x+3>x-5 =>A<0<=>x+3>0 và x-5<0

<=>-3<x<5

Vậy -3<x<5 thì A<0

10 tháng 11 2016

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

 

 

10 tháng 11 2016

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

14 tháng 8 2018

câu hỏi hay......nhưng tui xin nhường cho các bn khác

Hãy tích đúng cho tui nha

THANKS

4 tháng 8 2017

Với mọi n thuộc N* ta có :

\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)

\(=\left(n^2+n+\frac{1}{2}\right)\left(n^2-n+\frac{1}{2}\right)\)

\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)

\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)

\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)

26 tháng 8 2017

a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'

= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009

= 2010/2005

= 402/401

26 tháng 8 2017

\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)

\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)

\(=\frac{2010}{2005}\)

\(=\frac{402}{401}\)

Bài 1:...
Đọc tiếp

Bài 1: Tính

a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)

b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)

c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)

Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)

b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\) 

c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)

d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)

e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)

Bài 3: Chứng minh rằng

a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)

b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)

Bài 4: 

a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)

b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)

c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)

1

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)