Toán Lớp 8: tim x biet (2-x)^3+(3+x)(9-3x+x^2)+6x(1-x)=17 tim x biet (x+3)^2+(4-x)(x+4)=1 tim x biet x^4-2x^2+1=0

Question

Toán Lớp 8:  tim x biet (2-x)^3+(3+x)(9-3x+x^2)+6x(1-x)=17 tim x biet (x+3)^2+(4-x)(x+4)=1 tim x biet x^4-2x^2+1=0

haianhtu97 · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    tim x biet:      (2-x)^3+(3+x)(9-3x+x^2)+6x(1-x)=17     8-12x+6x^2-x^3+27+x^3+6x-6x^2=17     6x+23=17     6x=-6     x=-1     tim x biet:      (x+3)^2+(4-x)(x+4)=1     x^2+6x+9+16-x^2=1     6x+25=1     6x=-24     x=-4     tim x biet:      x^4-2x^2+1=0     (x^2-1)^2=0     x^2-1=0     x=&plusmn;1

ngochuong2k2 · Answer

a. $(2-x)^3+(3+x)(9-3x+x^2)+6x(1-x)=17$   $&hArr; 8-12x+6x^2-x^3+27+x^3+6x-6x^2=17$   $&hArr; -6x+35=17$   $&hArr; -6x=-18$   $&hArr; x=3$   b. $(x+3)^2+(4-x)(x+4)=1$   $&hArr; x^2+6x+9+16-x^2=1$   $&hArr; 6x+25=1$   $&hArr; 6x=-24$   $&hArr; x=-4$   c. $x^4-2x^2+1=0$   $&hArr; (x^2-1)^2=0$   $&hArr; x= pm 1$