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