Question

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

Giúp mừn vs mn ơy ^.^

Answer 1

ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0

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

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

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

Answer 2

*ĐK : 101x\(\ge\) 0 => x\(\ge\)0

=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)

     \(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)

...

\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

Ta có :

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

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

<=> \(100x+\frac{5050}{101}=101x\)

<=> \(100x-101x=\frac{-5050}{101}\)

<=> -x = -50

=> x = 50 ( t/m x >/ 0)