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6 tháng 5 2017

Ta có

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
\(\left|x+x+...x\right|+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\frac{99}{100}=100x\)
Sau đó tự biến đổi nha! Mik chỉ giải tới đó thôi vì mới lớp 6 à!

13 tháng 10 2018

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=100x\)

có :

\(\left|x+\frac{1}{1\cdot2}\right|;\left|x+\frac{1}{2\cdot3}\right|;\left|x+\frac{1}{3\cdot4}\right|;...;\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge\frac{0}{100}\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\)

\(=x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+x+\frac{1}{3\cdot4}+...+x+\frac{1}{99\cdot100}\)

bước này tự lm tp

đề chưa đầy đủ

19 tháng 3 2018

à đề thiếu tổng các giá trị tuyệt đối ở trên =100x

5 tháng 10 2018

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)

\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(50x=1-\frac{1}{100}\)

\(50x=\frac{99}{100}\)

\(x=\frac{99}{5000}\)

5 tháng 10 2018

Do \(\left|a\right|\ge0\forall a\) nên:

\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)

\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)

Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)

\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)

\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)

16 tháng 10 2018

các giá trị tuyệt đối trên có tổng lớn hơn hoặc bằng 0(>=0)

=>100x>=0

=>x>=0 =>x+1/(1.2) >0 ;x+1/(2.3)>0;x+1/(3.4);.....;x+1/(99.100)>0

=> ta có thể phá dấu giá trị tuyệt đối 

=>100x=x+x+...+x(có 99. x)+(1/(1.2)+1/(2.3)+..+1/(99.100))

=>100x=99x+99/100

=>x=99/100

4 tháng 11 2016

Ta có :

\(\begin{cases}\left|x+\frac{1}{1.2}\right|\ge0\\\left|x+\frac{1}{2.3}\right|\ge0\\...\\\left|x+\frac{1}{99.100}\right|\ge0\end{cases}\)\(\left(\forall x\right)\)

\(\Rightarrow100x>0\)

=> x > 0

=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+....+\left|x+\frac{1}{99.100}\right|\)

\(=x+\frac{1}{1.2}+x+\frac{1}{2.3}+.....+x+\frac{1}{99.100}=100x\)

\(\Rightarrow100x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=0\)

Dễ thấy VT \(\ne\)VP

=> \(x\in\varnothing\)

4 tháng 11 2016

Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;...;\left|x+\frac{1}{99.100}\right|\ge0\)

=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

=> \(100x\ge0\Rightarrow x\ge0\)

=> \(\left|x+\frac{1}{1.2}\right|=\left(x+\frac{1}{1.2}\right);\left|x+\frac{1}{2.3}\right|=\left(x+\frac{1}{2.3}\right);...;\left|x+\frac{1}{99.100}\right|=\left(x+\frac{1}{99.100}\right)\)=> \(\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+...+\left(x+\frac{1}{99.100}\right)=100x\)

=> 99x + \(\frac{99}{100}\) = 100x

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

10 tháng 7 2018

a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)

\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)

\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)

\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)

=> x = 4

b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)

\(\Rightarrow100x-99x=1-\frac{1}{100}\)

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

3 tháng 2 2019

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+99\right|=100x\)

\(\left|x+1\right|\ge0;\left|x+2\right|\ge0;...;\left|x+99\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+...+x+99=100x\)

\(\Rightarrow99x+1+2+3+...+99=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow-x=-4950\)

\(\Rightarrow x=4950\)

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{49\cdot50}\right|=50x\)

\(\left|x+\frac{1}{1\cdot2}\right|\ge0;\left|x+\frac{1}{2\cdot3}\right|\ge0;...;\left|x+\frac{1}{49\cdot50}\right|\ge0\)

\(\Rightarrow50x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{49\cdot50}\)

\(\Rightarrow49x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

tu lam 

4 tháng 2 2019

\(a;\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+..............+\left|x+99\right|=100x^{\left(1\right)}\)

Ta có \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+3\right|\ge0;.............;\left|x+99\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\).Từ (1) \(\Rightarrow x+1+x+2+x+3+..................+x+99=100x\)

\(\Rightarrow\left(x+x+x+........+x\right)+\left(1+2+3+..........+99\right)=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow x=4950\)(t/m đk x > =  0)

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+.........+\left|x+\frac{1}{49.50}\right|=50x^{(∗)}\)

\(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;............;\left|x+\frac{1}{49.50}\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow50x\ge0\Rightarrow x\ge0\)

Với x > = 0 .Từ (*) \(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+............+x+\frac{1}{49.50}=50x\)

\(\Rightarrow\left(x+x+x+.......+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49.50}\right)=50x\)

\(\Rightarrow49x+\left(1-\frac{1}{50}\right)=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

\(\Rightarrow x=\frac{49}{50}\)(t/m đk \(x\ge0\))

6 tháng 7 2016

Cho mình bổ sung tí : b = 100x nhé 

Kết quả hình ảnh cho thank you