Toán Lớp 8: a)x^4+2x^2-24 b)x^3+x^2+4 c)x^4+x^3+3x^2+2x+2 tìmx a)(x^2+1)^2-4x^2=0 b)x^2-4x+3=0 c)x^2-7x-30=0

Question

Toán Lớp 8:  a)x^4+2x^2-24 b)x^3+x^2+4 c)x^4+x^3+3x^2+2x+2 tìmx a)(x^2+1)^2-4x^2=0 b)x^2-4x+3=0 c)x^2-7x-30=0

dantam45 · Answer

Giải đáp + Lời giải và giải thích chi tiết: a)x^4+2x^2-24 Đặt x^2=t ta có: t^2+2t−24 =t^2−4t+6t−24 =t(t−4)+6(t−4) =(t+6)(t−4) =(x^2+6)(x^2−4) =(x^2+6)(x−2)(x+2) b)x^3+x^2+4 = x ^3 − x^ 2 + 2 x + 2 x ^2 − 2 x + 4 x^3+x^2+4 =x^3−x^2+2x+2x^2−2x+4 = x ( x ^2 − x + 2 ) + 2 ( x ^2 − x + 2 ) = ( x + 2 ) ( x ^2 − x + 2 ) c)x^4+x^3+3x^2+2x+2 =x^4+x^3+x^2+2x^2+2x+2 =x^2(x^2+x+1)+2(x^2+x+1) =(x^2+2)(x^2+x+1) Tìm x: a)(x^2+1)^2-4x^2=0 <=> (x^2+1)2−(2x)^2=0 <=>(x^2+1−2x)(x^2+1+2x)=0 <=>(x+1)^2(x−1)^2=0 <=>(x+1)^2(x−1)(x+1)=0 =>$ left[ begin{matrix} x+1=0 x-1=0 end{matrix} right.$ <=>$ left[ begin{matrix} x=-1 x=1 end{matrix} right.$ b)x^2-4x+3=0 <=>x^2-x-3x+3=0 <=> x^2(x-1)-3(x-1)=0 <=>(x-1)(x^2-3)=0 =>$ left[ begin{matrix} x^2-3=0 x-1=0 end{matrix} right.$ <=>$ left[ begin{matrix} x= sqrt{3} x=1 end{matrix} right.$ c)x^2-7x-30=0 <=>x^2-10x+3x-30=0 <=>x(x-10)+3(x-10)=0 <=>(x-10)(x+3)=0 =>$ left[ begin{matrix} x+3=0 x-10=0 end{matrix} right.$ <=>$ left[ begin{matrix} x=-3 x=10 end{matrix} right.$