Câu hỏi của Trần Hữu Ngọc Minh

Question

Giải hệ phương trình$\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\end{cases}}$

Kiệt Nguyễn · Accepted Answer

$\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}$$ĐK:x,y>0$$\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0$Vì x, y > 0 nên $x+2y+\frac{1}{y\sqrt{x}}>0$suy ra x - y = 0 hay x = yThay x = y vào (2), ta được: $\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3$$\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}$$\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}$$\Leftrightarrow\sqrt{x+3}.\sqrt{x}-\sqrt{x+3}-\sqrt{x}+1=0$$\Leftrightarrow\left(\sqrt{x+3}-1\right)\left(\sqrt{x}-1\right)=0$$\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\x=1\left(tmđk\right)\end{cases}}\Rightarrow x=y=1$Vậy hệ có một nghiệm duy nhất $\left(x;y\right)=\left(1;1\right)$Câu trả lời: $\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}$$ĐK:x,y>0$$\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0$Vì x, y > 0 nên $x+2y+\frac{1}{y\sqrt{x}}>0$suy ra x - y = 0 hay x = yThay x = y vào (2), ta được: $\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3$$\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}$$\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}$$\Leftrightarrow\sqrt{x+3}.\sqrt{x}-\sqrt{x+3}-\sqrt{x}+1=0$$\Leftrightarrow\left(\sqrt{x+3}-1\right)\left(\sqrt{x}-1\right)=0$$\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\x=1\left(tmđk\right)\end{cases}}\Rightarrow x=y=1$Vậy hệ có một nghiệm duy nhất $\left(x;y\right)=\left(1;1\right)$

Tran Le Khanh Linh · Answer

$\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}$ĐK: $\hept{\begin{cases}x>0\y>0\end{cases}}$và $\hept{\begin{cases}x+3\ge0\x^2+3x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\y>0\end{cases}}}$$\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x+y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0\Leftrightarrow x=y$do $x+2y+\frac{1}{y\sqrt{x}}>0\forall x,y>0$Thay y=x vào pt (2) ta được$\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}$$\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\Leftrightarrow\sqrt{x+3}\cdot\sqrt{x}-\sqrt{x+3}-\sqrt{x+1}=0$$\Leftrightarrow\left(\sqrt{x+1}-1\right)\left(\sqrt{x}-1\right)=0$$\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\sqrt{x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\left(loai\right)\x=1\left(tm\right)\end{cases}\Rightarrow}x=y=1}$Vậy hệ có nghiệm duy nhất (x;y)=(1;1)Câu trả lời: $\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}$ĐK: $\hept{\begin{cases}x>0\y>0\end{cases}}$và $\hept{\begin{cases}x+3\ge0\x^2+3x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\y>0\end{cases}}}$$\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x+y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0\Leftrightarrow x=y$do $x+2y+\frac{1}{y\sqrt{x}}>0\forall x,y>0$Thay y=x vào pt (2) ta được$\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}$$\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\Leftrightarrow\sqrt{x+3}\cdot\sqrt{x}-\sqrt{x+3}-\sqrt{x+1}=0$$\Leftrightarrow\left(\sqrt{x+1}-1\right)\left(\sqrt{x}-1\right)=0$$\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\sqrt{x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\left(loai\right)\x=1\left(tm\right)\end{cases}\Rightarrow}x=y=1}$Vậy hệ có nghiệm duy nhất (x;y)=(1;1)

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