Câu hỏi của Khánh Quỳnh Nguyễn

Question

Cho hai số dương x y, thỏa mãn $x^3+y^3=3xy-1$Tính $A=x^{2018}+y^{2019}$

Trần Thanh Phương · Accepted Answer

$x^3+y^3=3xy-1$$\Leftrightarrow x^3+y^3-3xy+1=0$$\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3xy-3x^2y-3xy^2+1=0$$\Leftrightarrow\left(x+y\right)^3+1-3xy\left(x+y+1\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1\right)-3xy\left(x+y+1\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)=0$$\Rightarrow\orbr{\begin{cases}x+y+1=0\x^2+y^2-xy-x-y+1=0\end{cases}}$$\Rightarrow\orbr{\begin{cases}x+y=-1\x^2+y^2-xy-x-y+1=0\end{cases}}$Mà x, y dương nên $x+y=-1$là vô líVậy $x^2+y^2-xy-x-y+1=0$Đến đây đợi tớ nghĩ tiếp :vCâu trả lời: $x^3+y^3=3xy-1$$\Leftrightarrow x^3+y^3-3xy+1=0$$\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3xy-3x^2y-3xy^2+1=0$$\Leftrightarrow\left(x+y\right)^3+1-3xy\left(x+y+1\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1\right)-3xy\left(x+y+1\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)=0$$\Leftrightarrow\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)=0$$\Rightarrow\orbr{\begin{cases}x+y+1=0\x^2+y^2-xy-x-y+1=0\end{cases}}$$\Rightarrow\orbr{\begin{cases}x+y=-1\x^2+y^2-xy-x-y+1=0\end{cases}}$Mà x, y dương nên $x+y=-1$là vô líVậy $x^2+y^2-xy-x-y+1=0$Đến đây đợi tớ nghĩ tiếp :v

Hoài Lâm · Answer

X3 + Y3 =3XY - 1=> X3 + Y3 + 3X2Y + 3XY2 - 3X2Y - 3XY2 - 3XY + 1 = 0=> $\subset X+Y\supset^3$+ 1 - 3XY$\subset X+Y+1\supset$= 0=> $\subset X+Y+1\supset.$$\subset\subset X+Y\supset^2-X-Y+1\supset$-3XY$\subset X+Y+1\supset=0$=>$\subset X+Y+1\supset.$$\subset X^2+Y^2+2XY-X-Y+1-3XY\supset$=0=> $\subset X+Y+1\supset.\subset X^2+Y^2-XY-X-Y+1$=0Vì X,Y > 0 =>X+Y+1 > 0 $\Rightarrow X^2+Y^2-XY-X-Y+1=0$$\Rightarrow2X^2+2Y^2-2XY-2X-2Y+2=0$$\Rightarrow X^2-2XY+Y^2+X^2-2X+1+Y^2-2Y+1=0$$\Rightarrow\subset X-Y\supset^2+\subset X-1\supset^2+\subset Y-1\supset^2=0$Vì $\subset X-Y\supset^2\ge;\subset X-1\supset^2\ge0;\subset Y-1\supset^2\ge0$$\Rightarrow\hept{\begin{cases}\subset X-Y\supset^2=0\\subset X-1\supset^2=0\\subset Y-1\supset^2=0\end{cases}}$$\Rightarrow\hept{\begin{cases}X-Y=0\X-1=0\Y-1=0\end{cases}}$$\Rightarrow X=Y=1$ $\Rightarrow A=1+1=2$Câu trả lời: X3 + Y3 =3XY - 1=> X3 + Y3 + 3X2Y + 3XY2 - 3X2Y - 3XY2 - 3XY + 1 = 0=> $\subset X+Y\supset^3$+ 1 - 3XY$\subset X+Y+1\supset$= 0=> $\subset X+Y+1\supset.$$\subset\subset X+Y\supset^2-X-Y+1\supset$-3XY$\subset X+Y+1\supset=0$=>$\subset X+Y+1\supset.$$\subset X^2+Y^2+2XY-X-Y+1-3XY\supset$=0=> $\subset X+Y+1\supset.\subset X^2+Y^2-XY-X-Y+1$=0Vì X,Y > 0 =>X+Y+1 > 0 $\Rightarrow X^2+Y^2-XY-X-Y+1=0$$\Rightarrow2X^2+2Y^2-2XY-2X-2Y+2=0$$\Rightarrow X^2-2XY+Y^2+X^2-2X+1+Y^2-2Y+1=0$$\Rightarrow\subset X-Y\supset^2+\subset X-1\supset^2+\subset Y-1\supset^2=0$Vì $\subset X-Y\supset^2\ge;\subset X-1\supset^2\ge0;\subset Y-1\supset^2\ge0$$\Rightarrow\hept{\begin{cases}\subset X-Y\supset^2=0\\subset X-1\supset^2=0\\subset Y-1\supset^2=0\end{cases}}$$\Rightarrow\hept{\begin{cases}X-Y=0\X-1=0\Y-1=0\end{cases}}$$\Rightarrow X=Y=1$ $\Rightarrow A=1+1=2$

Hoàng Sơn · Answer

$x^3+y^3=3xy-1\Leftrightarrow x^3+y^3+1-3xy=0\ $$\Leftrightarrow$$\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy=0$$\Leftrightarrow$$\left(x+y\right)^3+1^3-3xy\left(x+y+1\right)=0$$\Leftrightarrow$$\left(x+y+1\right)\left(\left(x+y\right)^2-x-y+1-3xy\right)=0$$\Leftrightarrow$$\left(x+y+1\right)\left(x^2+y^2+1^2-x-y-xy\right)=0$$\Leftrightarrow$$x^2+y^2+1^2-x-y-xy=0$( Vì x+y+1 (> bằng 1) )$\Leftrightarrow$$2x^2+2y^2+2.1^2-2x-2y-2xy=0$$\Leftrightarrow$$\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0$$\Leftrightarrow$$\hept{\begin{cases}x-y=0\x-1=0\y-1=0\end{cases}}$(Do bình phương của 1 số thì luôn ( > bằng ) 0$\Leftrightarrow$$\hept{\begin{cases}x=1\y=1\end{cases}}$Thay vào A, ta có:$A=x^{2018}+y^{2019}=1^{2018}+1^{2019}=1+1=2$( CTV mà ngu vl )Câu trả lời: $x^3+y^3=3xy-1\Leftrightarrow x^3+y^3+1-3xy=0\ $$\Leftrightarrow$$\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy=0$$\Leftrightarrow$$\left(x+y\right)^3+1^3-3xy\left(x+y+1\right)=0$$\Leftrightarrow$$\left(x+y+1\right)\left(\left(x+y\right)^2-x-y+1-3xy\right)=0$$\Leftrightarrow$$\left(x+y+1\right)\left(x^2+y^2+1^2-x-y-xy\right)=0$$\Leftrightarrow$$x^2+y^2+1^2-x-y-xy=0$( Vì x+y+1 (> bằng 1) )$\Leftrightarrow$$2x^2+2y^2+2.1^2-2x-2y-2xy=0$$\Leftrightarrow$$\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0$$\Leftrightarrow$$\hept{\begin{cases}x-y=0\x-1=0\y-1=0\end{cases}}$(Do bình phương của 1 số thì luôn ( > bằng ) 0$\Leftrightarrow$$\hept{\begin{cases}x=1\y=1\end{cases}}$Thay vào A, ta có:$A=x^{2018}+y^{2019}=1^{2018}+1^{2019}=1+1=2$( CTV mà ngu vl )

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