Kayasari Ryuunosuke Coordinator
21/07/2017 at 21:56-
1st case : \(x< 1\Leftrightarrow x-1< 0;2x-4< -2\)
\(\Rightarrow\left|x-1\right|=1-x;\left|2x-4\right|=4-2x\).We have :
\(1-x+4-2x=3\Leftrightarrow5-3x=3\Leftrightarrow3x=2\Leftrightarrow x=\dfrac{2}{3}\)(satisifed)
2nd case : \(1\le x< 2\Leftrightarrow x-1\ge0;2x-4< 0\)
\(\Rightarrow\left|x-1\right|=x-1;\left|2x-4\right|=4-2x\).We have :
\(x-1+4-2x=3\Leftrightarrow3-x=3\Leftrightarrow x=0\) (absurd)
3rd case : \(x\ge2\Rightarrow x-1\ge1;2x-4\ge0\)
\(\Rightarrow\left|x-1\right|=x-1;\left|2x-4\right|=2x-4\).We have :
\(x-1+2x-4=3\Leftrightarrow3x-5=3\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(satisfied)
So,\(S=\left\{\dfrac{2}{3};\dfrac{8}{3}\right\}\)
Selected by MathYouLike
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Aim Egst 21/07/2017 at 23:52
\(\left|x-1\right|+\left|2x-4\right|=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(2x-4\right)^2}=3\)
\(\Leftrightarrow\sqrt{x^2-2x+1}+\sqrt{4x^2-16x+16}=3\)
\(\Leftrightarrow\sqrt{x^2-2x+1}-\left(\dfrac{2}{3}x-\dfrac{1}{9}\right)+\sqrt{4x^2-16x+16}-\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1-\left(\dfrac{2}{3}x-\dfrac{1}{9}\right)^2}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{4x^2-16x+16-\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)^2}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)
\(\Leftrightarrow\dfrac{\dfrac{45x^2-150x+80}{81}}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{\dfrac{288x^2-960x+512}{81}}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)
\(\Leftrightarrow\dfrac{\dfrac{5\left(3x-8\right)\left(3x-2\right)}{81}}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{\dfrac{32\left(3x-8\right)\left(3x-2\right)}{81}}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)
\(\Leftrightarrow\dfrac{\left(3x-8\right)\left(3x-2\right)}{81}\left(\dfrac{5}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{32}{\sqrt{4x^2-16x+16}-\dfrac{2}{3}x+\dfrac{28}{9}}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-8=0\\3x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Kayasari Ryuunosuke selected this answer.