x(x+15−x)(x+x+15−x)=6(DK:x=−1) ⇔x+1x2(5−x)+(x+1)2x(5−x)2=(x+1)26(x+1)2 ⇔(x+1)x2(5−x)+x(5−x)2=6(x+1)2 ⇔−x4+5x3−11x2+13x−6=0 ⇔(x−1)(−x3+4x2−7x+6)=0 ⇔(x−1)(x−2)(−x2+2x−3)=0
<=> x=1 hoặc x=2
vì −x2+2x−3≥−2 với mọi x
x(x+15−x)(x+x+15−x)=6(DK:x=−1) ⇔x+1x2(5−x)+(x+1)2x(5−x)2=(x+1)26(x+1)2 ⇔(x+1)x2(5−x)+x(5−x)2=6(x+1)2 ⇔−x4+5x3−11x2+13x−6=0 ⇔(x−1)(−x3+4x2−7x+6)=0 ⇔(x−1)(x−2)(−x2+2x−3)=0
<=> x=1 hoặc x=2
vì −x2+2x−3≥−2 với mọi x