Question

Tìm x :

Answer 1

| x - 1/2 | - 1/15 . (-3)² = 0,4

| x - 1/2 | - 1/15 .  9      = 2/5

| x - 1/2 | - 3/5              = 2/5

| x - 1/2 |                      = 2/5 + 3/5 = 1

\(\Rightarrow x-\dfrac{1}{2}=\pm1\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

 

Answer 2

\(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{15}.\left(-3\right)^2=0,4.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|-\dfrac{1}{15}.9=\dfrac{2}{5}.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|-\dfrac{3}{5}=\dfrac{2}{5}.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=1.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1.\\x-\dfrac{1}{2}=-1.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}.\\x=-\dfrac{1}{2}.\end{matrix}\right.\)

Answer 3

c ) \(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{15}.\left(-3\right)^2=0,4\)

TH1 : \(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\) khi \(x-\dfrac{1}{2}\ge0\)

TH2 : \(\left|x-\dfrac{1}{2}\right|=-x+\dfrac{1}{2}\)  khi \(x-\dfrac{1}{2}< 0\)

Với TH1 , ta có 

\(x-\dfrac{1}{2}-\dfrac{1}{15}.\left(-3\right)^2=0,4\\ x-\dfrac{13}{30}.9=0,4\\ x-3,9=0,4\\ x=4,3\)

Với TH2 . ta có

\(-x+\dfrac{1}{2}-\dfrac{1}{15}.\left(-3\right)^2=0,4\\ -x+\dfrac{13}{30}.9=0,4\\ -x+3,9=0,4\\ =-x=-3,5\\ x=3,5\)

Vậy x = { 3,5 ; 4,3 }