Toán Lớp 8: TÌM X: x^2+4x=12 b) x^2-3x+1=0 Làm theo cách lp 8 nhé

Question

Toán Lớp 8:  TÌM X: x^2+4x=12 b) x^2-3x+1=0 Làm theo cách lp 8 nhé

tuyetnhungthu · Answer

a)x^2+4x=12 <=>x^2+4x+4=16 <=>(x+2)^2=16 <=>[(x+2=4),(x+2=-4):} <=>[(x=2),(x=-6):} Vậy S={2;-6}. b)x^2-3x+1=0 <=>x^2-2*x*3/2+9/4=5/4 <=>(x-3/2)^2=5/4 <=> ( left[ begin{array}{l}x- dfrac32= dfrac{- sqrt5}{2} x- dfrac32= dfrac{ sqrt5}{2} end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{- sqrt5+3}{2} x= dfrac{3+ sqrt5}{2} end{array} right. ) Vậy S={ frac{- sqrt5+3}{2}, frac{3+ sqrt5}{2}}.

hoaiphuong85 · Answer

a) x^2+4x=12 <=> x^2+4x-12=0 <=> x^2-2x+6x-12=0 <=> x(x-2)+6(x-2)=0 <=> (x+6)(x-2)=0 <=> [(x+6=0),(x-2=0):} <=> [(x=-6),(x=2):} Vậy x in {-6;2} b) x^2-3x+1=0 <=> x^2-2.x . 3/2+9/4+1-9/4=0 <=> (x-3/2)^2-5/4=0 <=> (x-3/2)^2=5/4 <=> ( left[ begin{array}{l}x- dfrac{3}{2}= dfrac{ sqrt{5}}{2} x- dfrac{3}{2}=- dfrac{ sqrt{5}}{2} end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{3+ sqrt{5}}{2} x= dfrac{3- sqrt{5}}{2} end{array} right. ) Vậy x in {(3+-sqrt5)/2}