|
|
|
PT(1) tuong duong $y^{2} + (y-3)x -4y+3=0 \Leftrightarrow (y-3)(y+x-1)=0$ theo dk thi y=3 (loai)
suy ra $ x+y-1=0$ suy ra $y=1-x$ thay vao (2) ta duoc:
$\sqrt[3]{x-2} + \sqrt{x+1} =3$ ta dat :
$\left.\begin{matrix} \\ \end{matrix}\right\} \sqrt[3]{x-2} =t$ va $\sqrt{x+1}=3-t$ dieu kien $t\leq 2 $
ta binh phuong hai ve cua pt(3) va pt(4) rui lay ve tru ve ta duoc ket qua t=1 thay vao pt(3) duoc x=3
cac ban tu lam nhe
|