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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: lx – $\frac{1}{2}$l – $\sqrt[]{\frac{1}{9}}$ = $\sqrt[]{\frac{1}{4}}$ (lx – $\frac{1}{2}$l là giá trị tuyệt đối của x – $\frac{1}{2}

Toán Lớp 7: lx – $\frac{1}{2}$l – $\sqrt[]{\frac{1}{9}}$ = $\sqrt[]{\frac{1}{4}}$
(lx – $\frac{1}{2}$l là giá trị tuyệt đối của x – $\frac{1}{2}$)

Comments ( 2 )

  1. |x – 1/2| – sqrt{1/9} = sqrt{1/4}
    |x – 1/2| – 1/3 = 1/2
    |x – 1/2| = 1/2 + 1/3
    |x – 1/2| = 3/6 + 2/6
    |x – 1/2| = 5/6
    Ta có:
    TH1: x – 1/2 = 5/6
            x = 5/6 + 1/2
            x = 5/6 + 3/6
            x = 8/6
            x = 4/3
    TH2: x – 1/2 = frac{-5}{6}
            x = frac{-5}{6} + 1/2
            x = frac{-5}{6} + 3/6 
            x = frac{-2}{6}
            x = frac{-1}{3}
    Vậy x = 4/3 hoặc x = frac{-1}{3}

  2. Answer
    |x – 1/2| – \sqrt{1/9} = \sqrt{1/4}
    => |x – 1/2| – 1/3 = 1/2
    => |x – 1/2| = 1/2 + 1/3
    => |x – 1/2| = 5/6
    => $\left[\begin{matrix} x – \dfrac{1}{2} = \dfrac{5}{6}\\ x – \dfrac{1}{2} = \dfrac{-5}{6}\end{matrix}\right.$
    => $\left[\begin{matrix} x = \dfrac{5}{6} + \dfrac{1}{2}\\ x = \dfrac{-5}{6} + \dfrac{1}{2}\end{matrix}\right.$
    => $\left[\begin{matrix} x = \dfrac{4}{3}\\ x = \dfrac{-1}{3}\end{matrix}\right.$
    $\text{Vậy}$ x \in {4/3 ; -1/3}

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222-9+11+12:2*14+14 = ? ( )