2 . ( `x^2` – 2x ) $^{2}$ = 3`x^2` – 6x + 9 Giải PT 26/08/2021 Bởi Athena 2 . ( `x^2` – 2x ) $^{2}$ = 3`x^2` – 6x + 9 Giải PT
Cậu tham khảo nhé: 2.($x^{4}$-4$x^{3}$+4$x^{2}$ )=3.$x^{2}$+6x+9 ⇔$2x^{4}$-8$x^{3}$+5$x^{2}$+6x-9=0 ⇔$2x^{4}$-$6x^{3}$-$2x^{3}$+$6x^{2}$-$x^{2}$+3x+3x-9=0 ⇔$2x^{3}$(x-3)-$2x^{2}$(x-3)-x(x-3)+3(x-3)=0 ⇔($2x^{3}$-$2x^{2}$-x+3).(x+3)=0 ⇔($2x^{3}$+$2x^{2}$-$4x^{2}$+4x+3x+3)(x-3)=0 ⇔($2x^{2}$-4x+3)(x+1)(x-3)=0 ⇔$\left \{ {{x=-1} \atop {x=3}} \right.$ ($2x^{2}$-4x+3)>0 cậu tự cm nhé Bình luận
Cậu tham khảo nhé:
2.($x^{4}$-4$x^{3}$+4$x^{2}$ )=3.$x^{2}$+6x+9
⇔$2x^{4}$-8$x^{3}$+5$x^{2}$+6x-9=0
⇔$2x^{4}$-$6x^{3}$-$2x^{3}$+$6x^{2}$-$x^{2}$+3x+3x-9=0
⇔$2x^{3}$(x-3)-$2x^{2}$(x-3)-x(x-3)+3(x-3)=0
⇔($2x^{3}$-$2x^{2}$-x+3).(x+3)=0
⇔($2x^{3}$+$2x^{2}$-$4x^{2}$+4x+3x+3)(x-3)=0
⇔($2x^{2}$-4x+3)(x+1)(x-3)=0
⇔$\left \{ {{x=-1} \atop {x=3}} \right.$ ($2x^{2}$-4x+3)>0 cậu tự cm nhé