Điểm cao có thưởng: -10x^4+83x^3-275x^2+412x-192=0 07/07/2021 Bởi Kennedy Điểm cao có thưởng: -10x^4+83x^3-275x^2+412x-192=0
$-10.x^4+83.x³-275.x²+412x-192= 0$ ⇔ $-10.x^4+30.x³+53x³-159x²-116x²+348x+64x-192= 0$ ⇔ $-10.x³.( x-3)+53.x².( x-3)-116x.( x-3)+64.( x-3)= 0$ ⇔ $( x-3).( -10.x³+53.x²-116x+64)= 0$ ⇔ $( x-3).( -10x³+8x²+45x²-36x-80x+64)= 0$ ⇔ $( x-3).[ -10x².( x-0,8)+45x.( x-0,8)-80.(x-0,8)]= 0$ ⇔ $( x-3).( x-0,8).( -10x²+45x-80)= 0$ Vì $-10x²+45x-80$ $=-10.(x²-4,5x+8)$ $=-10.(x²-2.2,25.x+5,0625+2,9375)$ $=-10.[(x-2,25)²+2,9375]$≠$0$ ⇒ $( x-3).( x-0,8)= 0$ ⇔ $x= 3$ hoặc $x= 0,8$ Bình luận
$-10.x^4+83.x³-275.x²+412x-192= 0$
⇔ $-10.x^4+30.x³+53x³-159x²-116x²+348x+64x-192= 0$
⇔ $-10.x³.( x-3)+53.x².( x-3)-116x.( x-3)+64.( x-3)= 0$
⇔ $( x-3).( -10.x³+53.x²-116x+64)= 0$
⇔ $( x-3).( -10x³+8x²+45x²-36x-80x+64)= 0$
⇔ $( x-3).[ -10x².( x-0,8)+45x.( x-0,8)-80.(x-0,8)]= 0$
⇔ $( x-3).( x-0,8).( -10x²+45x-80)= 0$
Vì $-10x²+45x-80$
$=-10.(x²-4,5x+8)$
$=-10.(x²-2.2,25.x+5,0625+2,9375)$
$=-10.[(x-2,25)²+2,9375]$≠$0$
⇒ $( x-3).( x-0,8)= 0$
⇔ $x= 3$ hoặc $x= 0,8$