Question

Bài 1:Tìm x, biết:

1) | \(\text{|}x+\dfrac{1}{2}\text{|}+\text{|}x+\dfrac{1}{3}\text{|}+\text{|}x+\dfrac{1}{6}\text{|}=4x\)

Answer 1

\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)

Ta có:

\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)

\(\Rightarrow3x+1=4x\)

\(\Rightarrow x=1\)

Answer 2

Với mọi giá trị của \(x\in R\) ta có:

\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)

Thay vào ta được:

\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)

\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)

Vậy...................

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