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

Toán Lớp 7: tìm x `(x-2)^2021=(x-2)^2020` `|x+1|+|x+2|+|x+3|+|x+4|=5x-20`

Toán Lớp 7: tìm x
(x-2)^2021=(x-2)^2020
|x+1|+|x+2|+|x+3|+|x+4|=5x-20

Comments ( 1 )

  1. a)
    (x-2)^2021 = (x-2)^2020
    => (x-2)^2021 – (x-2)^2020 = 0
    => (x-2)^2020 . [ (x-2) – 1] =0
    => (x-2)^2020 . (x-3) =0
    => (x-2)^2020=0 hoặc x-3=0
    +) (x-2)^2020 = 0
    => x – 2 = 0
    => x=2
    +) x- 3=0
    =>x=3
    Vậy x \in {2;3}
    b)
    |x+1| + |x+2| + |x+3| + |x+4| = 5x-20 (1)
    \forall x ta có :
    |x+1| \ge 0
    |x+2| \ge0
    |x+3| \ge0
    |x+4| \ge0
    => |x+1| + |x+2| + |x+3| + |x+4| \ge 0
    Mà |x+1| + |x+2| + |x+3| + |x+4| = 5x-20 nên 5x – 20 \ge 0
    => 5x \ge 20
    => x \ge 4
    Với x\ge 4 thì (1) trở thành :
    (x+1) + (x+2)+ (x+3) +(x+4) = 5x – 20
    => x + 1 + x + 2 + x + 3 + x  + 4 = 5x-20
    => 4x + 10 = 5x – 20
    => 5x – 4x = 10 + 20
    =>x=30 (thỏa mãn điều kiện)
    Vậy x=30

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

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