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5 tháng 10 2018

\(\left|x+\frac{1}{1\cdot5}\right|+\left|x+\frac{1}{5\cdot9}\right|+...+\left|x+\frac{1}{397\cdot401}\right|=101x\left(1\right)\)

Điều kiện:\(101x\ge0\)\(\Rightarrow\left|x+\frac{1}{1\cdot5}\right|\ge0;\left|x+\frac{1}{5\cdot9}\right|\ge0;.....;\left|x+\frac{1}{397\cdot401}\right|\ge0\)

Do vậy\(\left(1\right)\)trở thành:\(x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=101x\)

\(\left(x+x+x+..+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+..+\frac{1}{397\cdot401}\right)\)

Có 100 số x

\(\Leftrightarrow\)\(100x+\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}\right)=101x\)

\(\Leftrightarrow\)\(100x+\frac{1}{4}\left(1-\frac{1}{401}\right)=101x\)

\(\Leftrightarrow100x+\frac{1}{4}\left(\frac{400}{401}\right)=101x\)

\(\Leftrightarrow\)\(x=\frac{1}{4}\cdot\frac{400}{401}\)\(=\frac{100}{401}\)

30 tháng 10 2017

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)

17 tháng 2 2017

Ta có: \(\left|x+\frac{1}{101}\right|\ge0\); \(\left|x+\frac{2}{101}\right|\) \(\ge0\); ...; \(\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\); \(\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\); ...; \(\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

Thay vào đề bài ta đc:

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Rightarrow\) \(100x\) + \(\left(\frac{1+2+...+101}{101}\right)=101x\)

\(\Rightarrow100x+101=101x\)

\(\Rightarrow x=101\)

Vậy \(x=101.\)

17 tháng 2 2017

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x (1)

điều kiện:101x\(\ge\) 0 \(\Rightarrow\) x\(\ge\) 0

từ (1) \(\Rightarrow\) \(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}\)=101x

\(\Rightarrow\) 100x+(\(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\))=101x

\(\Rightarrow\) 100x+\(\frac{5050}{101}\)=101x

\(\Rightarrow\) \(\frac{5050}{101}\)=101x-100x

\(\Rightarrow\) x=50

k bt mk lm sai hay lm đúng nữa

nếu mk lm sai thì thôi nha!

22 tháng 10 2018

Vì \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(100x+\frac{1+2+...+100}{101}=101x\)

\(101x-100x=\frac{5050}{101}\)

\(x=50\)

Vậy x = 50

22 tháng 10 2018

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)

\(KĐ:101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)

\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)

\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)

\(x=\frac{101.100:2}{101}\)

\(x=50\)

a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1  - \(\frac{1}{4}\)\(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{x}{3}\)\(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)

Với \(\frac{x}{3}\) = \(\frac{33}{100}\)

\(\Rightarrow\)100x= 33.3

 \(\Rightarrow\)100x=99

\(\Rightarrow\)x=\(\frac{99}{100}\)

Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)

\(\Rightarrow\)100x=-33.3

\(\Rightarrow\)100x=-99

\(\Rightarrow\)x=\(\frac{-99}{100}\)

Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)

b, \(\frac{4}{1.5}\)\(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)

\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)

Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)

\(\Rightarrow\)(5x-4).101=100.101

\(\Rightarrow\)505x-404=10100

\(\Rightarrow\)505x=10504

\(\Rightarrow\)x=\(\frac{104}{5}\)

Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)

\(\Rightarrow\)(5x-4). 101=-100.101

\(\Rightarrow\)505x-404=-10100

\(\Rightarrow\)505x=-9696

\(\Rightarrow\)x=\(\frac{-96}{5}\)

Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)

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

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

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

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

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

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

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

13 tháng 9 2016

Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{1}{102}\right|+....+\left|x+\frac{100}{101}\right|>0\)

\(\Rightarrow101x>0\)

\(\Rightarrow x>0\)

\(\Rightarrow\left(x+\frac{1}{101}\right)+.....+\left(x+\frac{100}{101}\right)=101x\)

\(\Rightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow x=\frac{\left(100+1\right)100:2}{101}\)

\(\Rightarrow x=\frac{50.101}{101}\)

\(\Rightarrow x=50\)

Vậy x = 50

13 tháng 9 2016

Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)

=> \(101x\ge0\)

=> \(x\ge0\)

=> \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

            100 số x                          100 phân số

=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

11 tháng 10 2018

Do \(\left|a\right|\ge0\) nên:

a) \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\ge0\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\) (100 số hạng x)

\(\Leftrightarrow100x+5050=101x\Leftrightarrow201x=5050\Leftrightarrow x=\frac{5050}{201}\)

b) Đề sai nhé!

11 tháng 10 2018

Chết,nhầm ở câu cuối cùng của câu a) . Mình là ẩu thật :v. Sửa lại nhé:

\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow100x+50=101x\Leftrightarrow201x=50\Leftrightarrow x=\frac{50}{201}\)

19 tháng 9 2016

a ) \(3-4.\left|5-6x\right|=7\)

\(\Leftrightarrow4.\left|5-6x\right|=-4\)

\(\Leftrightarrow\left|5-6x\right|=-1\)

\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )

19 tháng 9 2016

b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)

=> \(4x\ge0\)

=> \(x\ge0\)

Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)

=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)

=> \(3x+\frac{31}{10}=4x\)

=> \(4x-3x=\frac{31}{10}\)

=> \(x=\frac{31}{10}\)

Vậy \(x=\frac{31}{10}\)

c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)

=> \(101x\ge0\)

=> \(x\ge0\)

Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

               100 số x

=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

Vậy x = 50