Tìm x,y,z : 2(x – 4)^10+ 8(y – 5)^20 +|z^3 + 27|=0 02/07/2021 Bởi Savannah Tìm x,y,z : 2(x – 4)^10+ 8(y – 5)^20 +|z^3 + 27|=0
$2(x – 4)^{10}+ 8(y – 5)^{20} +|z^3 + 27|=0$ Vì: $2(x-4)^{10}≥0∀x;8(y-5)^{20}≥0∀y;|z^3+27|≥0∀z$ Dấu “=” xảy ra khi $x-4=0;y-5=0;z^3+27=0$ $⇒x=4;y=5;z=-3$ Vậy $x=4;y=5;z=-3$ Bình luận
Đó nha bạn
$2(x – 4)^{10}+ 8(y – 5)^{20} +|z^3 + 27|=0$
Vì: $2(x-4)^{10}≥0∀x;8(y-5)^{20}≥0∀y;|z^3+27|≥0∀z$
Dấu “=” xảy ra khi $x-4=0;y-5=0;z^3+27=0$
$⇒x=4;y=5;z=-3$
Vậy $x=4;y=5;z=-3$