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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 taduoc ket qua t=1 thay vao pt(3) duoc x=3 cac ban tu lam nhe

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

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 taduoc ket qua t=1 thay vao pt(3) duoc x=3 cac ban tu lam nhe

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 taduoc ket qua t=1 thay vao pt(3) duoc x=3 cac ban tu lam nhe