Question

Bài 1: Tìm x, biết:

1) \(\text{|}x+1\text{|}+\text{|}x+2\text{|}+\text{|}x+3\text{|}+...+\text{|}x+19\text{|}=20x\)

Answer 1

\( \left| x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+19\right|=20x\)

\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\..................\\\left|x+19\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+19\right|\ge0\)

\(\Rightarrow20x\ge0\)

\(\Rightarrow x+1+x+2+...+x+19=20x\)

\(\Rightarrow19x+\left(1+2+...+19\right)=20x\)

\(\Rightarrow19x+190=20x\)

\(\Rightarrow x=190\)

Answer 2

Với mọi giá trị của \(x\in R\) ta có:

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+19\right|\ge0\)

\(\Rightarrow20x\ge0\Rightarrow x\ge0\)

\(\left\{{}\begin{matrix}x+1>0\\x+2>0\\..........\\x+19>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\...........\\\left|x+19\right|=x+19\end{matrix}\right.\)

Thay vào ta có:

\(x+1+x+2+x+3+...+x+19=20x\)

\(\Rightarrow19x+\left(1+2+3+....+19\right)=20x\)

\(\Rightarrow x=\dfrac{\left(1+19\right).19}{2}=190\)

Vậy..................

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Answer 3

x = 190