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

Toán Lớp 7: tìm x y z biết |3x-5|+[2y+5]^208+[4z-3]^20<0

Tháng Sáu 11, 2022 Toán học 2 Comments 0 views

Toán Lớp 7: tìm x y z biết |3x-5|+[2y+5]^208+[4z-3]^20<0

Comments ( 2 )

  1. kimoanh34
    kimoanh34
    11 Tháng Sáu, 2022 at 1:29 chiều
    Reply
    Answer
    $\text{Ta có:}$
    |3x – 5| \le 0 \forall x \in QQ
    |2y + 5|^{208} \le 0 \forall y \in QQ
    |4z – 3|^{20} \le 0 \forall z \in QQ
    => |3x – 5| + |2y + 5|^{208} + |4z – 3|^{20} \le 0 \forall x,y,z \in QQ
    => |3x – 5| + |2y + 5|^{208} + |4z – 3|^{20} = 0
    => {(3x-5=0),(2y+5=0),(4z-3=0):}
    => {(3x=0+5),(2y=0-5),(4z=0+3):}
    => {(3x=5),(2y=-5),(4z=3):}
    => {(x=5:3),(y=-5:2),(z=3:4):}
    => {(x=5/3),(y=-5/2),(z=3/4):}
    $\text{Vậy}$ (x , y , z) = (5/3 , -5/2 , 3/4)

  2. ngocle44
    ngocle44
    11 Tháng Sáu, 2022 at 1:29 chiều
    Reply
    Ta có :
    |3x-5|+(2y+5)^208+(4z-3)^20\ge0
    Mà |3x-5|\ge0; (2y-5)^208\ge0;(4z-3)\ge0
    Suy ra : |3x-5|+(2y+5)^208+(4z-3)^20=0
    =>$\begin{cases}3x-5=0\\2y+5=0\\4z-3=0 \end{cases}$
    =>$\begin{cases} x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4} \end{cases}$
    Vậy x=5/3; y=-5/2; z=3/4
     

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

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