Toán Lớp 8: tìm giá trị x, biết: `x^2 = 2x +1`

Question

Toán Lớp 8:  tìm giá trị x, biết: x^2 = 2x +1

truonganhthi · Answer

x^2=2x+1 <=>x^2-2x-1=0 <=>x^2-2x+1-2=0 <=>(x^2-2x+1)-2=0 <=>(x-1)^2=2 <=>(x-1)=+- sqrt{2} <=>$ left[ begin{matrix} x-1= sqrt{2} x-1=- sqrt{2} end{matrix} right.$ <=>$ left[ begin{matrix} x= sqrt{2}+1 x=- sqrt{2}+1 end{matrix} right.$ =>$ left[ begin{matrix} x=1+ sqrt{2} x=1- sqrt{2} end{matrix} right.$ Vậy S={1+ sqrt{2}; 1- sqrt{2}}

tonhu12 · Answer

Theo mik đề sai nha                  &darr;         x^2 = 2x - 1        x^2 - 2x + 1 = 0       ( x - 1 )^2     = 0         x - 1            = 0   => x               = 1           Nếu đề đ&uacute;ng th&igrave;                    &darr;          x^2 = 2x + 1          x^2 - 2x - 1 = 0          x^2 - 2x + 1 - 2 = 0         ( x - 1 )^2           = 2   =>  ( left[  begin{array}{l}x-1= sqrt{2}  x-1=- sqrt{2} end{array}  right. )    =>  ( left[  begin{array}{l}x= sqrt{2} + 1  x=- sqrt{2} + 1 end{array}  right. )