Hệ phương trình:::::::::::

N

nguyenbahiep1

từ hệ (1)

[laTEX]x = 2y^2 + y +1[/laTEX]

thay vào (2)

[laTEX]\sqrt{2y^2+2y+1}+\sqrt{2y^2-y+1} = 3y [/laTEX]

ta có

[laTEX]2y^2+2y+1-(2y^2-y+1) =\sqrt{2y^2+2y+1}^2-\sqrt{2y^2-y+1}^2 = 3y [/laTEX]

vậy

[laTEX]\sqrt{2y^2+2y+1}+\sqrt{2y^2-y+1} = (\sqrt{2y^2+2y+1}+\sqrt{2y^2-y+1})(\sqrt{2y^2+2y+1}-\sqrt{2y^2-y+1}) \\ \\ \Rightarrow \sqrt{2y^2+2y+1}-\sqrt{2y^2-y+1} = 1 \\ \\ \sqrt{2y^2+2y+1} = 1+\sqrt{2y^2-y+1} \\ \\ 2y^2+2y+1 = 1+ 2y^2-y+1 + 2.\sqrt{2y^2-y+1} \\ \\ 3y-1 = 2\sqrt{2y^2-y+1} \\ \\ y = 3 \Rightarrow x = 22[/laTEX]
 
You must log in or register to reply here.
Top Bottom