Toán Lớp 8: Giải pt (X^2-3x+3)(x^2-2x+3)=2x^2 Nhanh nha

Question

Toán Lớp 8:  Giải pt  (X^2-3x+3)(x^2-2x+3)=2x^2 Nhanh nha

annhocbichchaubich · Answer

Giải đáp + giải th&iacute;ch c&aacute;c bước giải:     (x^2-3x+3)(x^2-2x+3)=2x^2   Đặt x^2-2x+3=t, phương tr&igrave;nh c&oacute; dạng:   (t-x)t=2x^2   ->t^2-xt-2x^2=0   ->t^2+xt-2xt-2x^2=0   ->t(t+x)-2x(t+x)=0   ->(t-2x)(t+x)=0   -> ( left[  begin{array}{l}t-2x=0  t+x=0 end{array}  right. )    -> ( left[  begin{array}{l}x^2-2x+3-2x=0  x^2-2x+3+x=0 end{array}  right. )    -> ( left[  begin{array}{l}x^2-4x+3=0  x^2-x+3=0 end{array}  right. )    -> ( left[  begin{array}{l}x^2-x-3x+3=0  x^2-2.x. dfrac{1}{2}+ dfrac{1}{4}+ dfrac{11}{4}=0 end{array}  right. )    -> ( left[  begin{array}{l}x(x-1)-3(x-1)=0  (x- dfrac{1}{2})^2+ dfrac{11}{4}=0  text{ (v&ocirc; l&yacute;)} end{array}  right. )    ->(x-3)(x-1)=0   -> ( left[  begin{array}{l}x-3=0  x-1=0 end{array}  right. )    -> ( left[  begin{array}{l}x=3  x=1 end{array}  right. )