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

Toán Lớp 7: `|x+1/3|+4/5=|(-3,2)+2/5|+(1/49-1/3^2)(1/49-1/4^2)…(1/49-1/2018^2)` TÌM X

Toán Lớp 7: |x+1/3|+4/5=|(-3,2)+2/5|+(1/49-1/3^2)(1/49-1/4^2)…(1/49-1/2018^2)
TÌM X

Comments ( 2 )

  1. | x + 1/3 | + 4/5 = | ( – 3,2 ) + 2/5 | + ( 1/49 – 1/3^2 ) . ( 1/49 – 1/4^2 ) …. ( 1/49 – 1/2018^2 )
    ⇔ | x + 1/3 | + 4/5 = | ( – 16 )/5 + 2/5 | + ( 1/49 – 1/3^2 ) . ( 1/49 – 1/4^2 ) …. ( 1/49 – 1/7^2 ) …. ( 1/49 – 1/2018^2 )
    ⇔ | x + 1/3 | + 4/5 = | – 14/5 | + ( 1/49 – 1/3^2 ) . ( 1/49 – 1/4^2 ) …. 0 …. ( 1/49 – 1/2018^2 )
    ⇔ | x + 1/3 | + 4/5 = 14/5 + 0
    ⇔ | x + 1/3 | + 4/5 = 14/5
    ⇔ | x + 1/3 | = 14/5 – 4/5
    ⇔ | x + 1/3 | = 2
    ⇔ \(\left[ \begin{array}{l}x + \frac{1}{3} = 2\\x + \frac{1}{3} = – 2\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x = \frac{5}{3}\\x = \frac{- 7}{3}\end{array} \right.\) 
    Vậy , x ∈ { 5/3 ; ( – 7 )/3 } .

  2. Answer
    |x + 1/3| + 4/5 = |(-3,2) + 2/5| + (1/49 – 1/3^{2}) . (1/49 – 1/4^{2}) … (1/49 – 1/7^{2}) … (1/49 – 1/2018^{2})
    $\Rightarrow$ |x + 1/3| + 4/5 = |-16/5 + 2/5| + (1/49 – 1/3^{2}) . (1/49 – 1/4^{2}) … (1/49 – 1/49) … (1/49 – 1/2018^{2})
    $\Rightarrow$ |x + 1/3| + 4/5 = |-16/5 + 2/5| + (1/49 – 1/3^{2}) . (1/49 – 1/4^{2}) … 0 … (1/49 – 1/2018^{2})
    $\Rightarrow$ |x + 1/3| + 4/5 = |-16/5 + 2/5| + 0
    $\Rightarrow$ |x + 1/3| + 4/5 = |-16/5 + 2/5|
    $\Rightarrow$ |x + 1/3| + 4/5 = |-14/5|
    $\Rightarrow$ |x + 1/3| + 4/5 = 14/5
    $\Rightarrow$ |x + 1/3| = 14/5 – 4/5
    $\Rightarrow$ |x + 1/3| = 2
    $\Rightarrow$ $\left[\begin{matrix} x+\dfrac{1}{3}=2\\ x+\dfrac{1}{3}=-2\end{matrix}\right.$ 
    $\Rightarrow$ $\left[\begin{matrix} x=2-\dfrac{1}{3}\\ x=-2-\dfrac{1}{3}\end{matrix}\right.$ 
    $\Rightarrow$ $\left[\begin{matrix} x=\dfrac{5}{3}\\ x=\dfrac{-7}{3}\end{matrix}\right.$ 
    $\text{Vậy}$ x \in {5/3 ; -7/3}

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