Toán Lớp 7: Tìm x, y, z thỏa mãn: (3x-5)^2020 + (y^2-1)^2022 + (x-z)^2024 = 0
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Toán Lớp 7: Tìm x, y, z thỏa mãn: (3x-5)^2020 + (y^2-1)^2022 + (x-z)^2024 = 0
Comments ( 2 )
~ Bạn tham khảo ~
Ta có : (3x-5)^2020 >=0;(y^2-1)^2022>=0;(x-z)^2024>=0
=> (3x-5)^2020+(y^2-1)^2022+(x-z)^2024>=0
Mà (3x-5)^2020+(y^2-1)^2022+(x-z)^2024=0
$⇒\begin{cases} 3x-5=0\\y^2-1=0\\x-z=0 \end{cases}$
$⇔\begin{cases} x=\dfrac53\\y=\pm 1\\x=z \end{cases}$
$⇔\begin{cases} x=\dfrac53\\y=\pm 1\\z=\dfrac53 \end{cases}$
Vậy (x;y;z)={(5/3;1;5/3);(5/3;-1;5/3)}
Giải đáp:
x=5/3;y=+-1;z=5/3
Lời giải và giải thích chi tiết:
(3x-5)^2020+(y^2-1)^2022+(x-z)^2024=0
Ta có:
{((3x-5)^2020ge0forallx),((y^2-1)^2022ge0forally),((x-z)^2024ge0forallx;z):}
=>(3x-5)^2020+(y^2-1)^2022+(x-z)^2024ge0forallx;y;z
Mà (3x-5)^2020+(y^2-1)^2022+(x-z)^2024=0
=>{((3x-5)^2020=0),((y^2-1)^2022=0),((x-z)^2024=0):}
=>{(3x-5=0),(y^2-1=0),(x-z=0):}
=>{(3x=5),(y^2=1),(x=z):}
=>{(x=5/3),(y=+-1),(z=5/3):}
Vậy x=5/3;y=+-1;z=5/3