Toán Lớp 8: Tìm x x^2(x-2021)-x+2021=0 2x^2-3x-5

Question

Toán Lớp 8:  Tìm x x^2(x-2021)-x+2021=0 2x^2-3x-5

maianhnhiquynh · Answer

x^2(x-2021)-x+2021=0 <=>x^2(x-2021)-(x-2021)=0 <=>(x^2-1)(x-2021)=0 <=>(x+1)(x-1)(x-2021)=0 <=>[(x+1=0),(x-1=0),(x-2021=0):} <=>[(x=-1),(x=1),(x=2021):} Vậy x in{-1; 1; 2021} Sửa đề thành: 2x^2-3x-5=0 <=>2x^2+2x-5x-5=0 <=>(2x^2+2x)-(5x+5)=0 <=>2x(x+1)-5(x+1)=0 <=>(2x-5)(x+1)=0 <=>[(2x-5=0),(x+1=0):} <=>[(2x=5),(x=-1):} <=>$ left[ begin{matrix} x= dfrac{5}{2} x=-1 end{matrix} right.$ Vậy x=5/2 hoặc x=-1

thudan · Answer

a) x^2(x-2021) - x+2021 = 0 <=> x^2(x-2021) - (x-2021) = 0 <=> (x^2 - 1)(x-2021) = 0 <=> ( left[ begin{array}{l}x^2 - 1 =0 x-2021 =0 end{array} right. ) <=> ( left[ begin{array}{l}x^2 = 1 x=2021 end{array} right. ) <=> ( left[ begin{array}{l}x= ±1 x=2021 end{array} right. ) Vậy x∈{1;-1;2021} b) 2x^2 - 3x - 5 = 0 <=> 2x^2 + 2x - 5x - 5 =0 <=> 2x(x+1) - 5(x+1) = 0 <=> (2x-5)(x+1) =0 <=> ( left[ begin{array}{l}2x-5=0 x+1 =0 end{array} right. ) <=> ( left[ begin{array}{l}x=2,5 x=-1 end{array} right. ) Vậy x∈{2,5;-1}