(x^2-5).(x+2)+5x=2x^2+17 mn ơi giúp mik vs ạ 26/07/2021 Bởi Adalynn (x^2-5).(x+2)+5x=2x^2+17 mn ơi giúp mik vs ạ
$(x^2-5).(x+2)+5x=2x^2+17$ $x^3+2x^2-5x-10+5x=2x^2+17$ $x^3+2x^2-5x-10+5x-2x^2-17=0$ $x^3-27=0$ $x^3=27$ $x=3$ Vậy $x=3$ Bình luận
(x2 – 5)(x + 2) + 5x = 2x2 + 17 x3 + 2x2 – 5x – 10 + 5x = 2x2 + 17 x3 + 2x2 – 10 = 2x2 + 17 x3 – 10 = 17 x3 = 17 + 10 x3 = 27 x3 = 33 x = 3 Bình luận
$(x^2-5).(x+2)+5x=2x^2+17$
$x^3+2x^2-5x-10+5x=2x^2+17$
$x^3+2x^2-5x-10+5x-2x^2-17=0$
$x^3-27=0$
$x^3=27$
$x=3$
Vậy $x=3$
(x2 – 5)(x + 2) + 5x = 2x2 + 17
x3 + 2x2 – 5x – 10 + 5x = 2x2 + 17
x3 + 2x2 – 10 = 2x2 + 17
x3 – 10 = 17
x3 = 17 + 10
x3 = 27
x3 = 33
x = 3