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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
=> \(x\in\varnothing\)
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 = \(\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(1,2\div0,8=\left(-3,6\right)\div\left(3x\right)\)
\(1,5=\left(-3,6\right)\div3x\)
\(3x=\left(-3,6\right):1,5\)
\(3x=-2,4\)
\(x=-0,8\)
\(1,2\div8=\left(-3,6\right)\div3x\)
\(\Leftrightarrow\frac{3}{20}=\frac{-12}{x}\)
\(\Rightarrow3x=-240\)
\(\Rightarrow x=-80\)
~Std well~
#Tử_Trân