giup vs

toán lớp 9 hử bn

\(\hept{\begin{cases}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9-8y=-52-4xy\end{cases}\left(ĐK\hept{\begin{cases}x\in R\\y\ge-1\end{cases}}\right)}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\\left(x^3-4x^2\sqrt{y+1}+4xy+4x\right)-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\x\left(x-2\sqrt{y+1}\right)^2-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\x\left(3+2\sqrt{y+1}-2\sqrt{y+1}\right)^2-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\9x-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-4x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-x-2y+13=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-2\sqrt{y+1}-2y+10=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\\sqrt{y+1}=y-5\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow y+1=y^2-10y+25\left(y\ge1\right)\)

\(\Leftrightarrow y^2-11y+24=0\)

\(\Delta=11^2-4.24=25>0\)

\(\Rightarrow\)pt (1) có 2 nghiệm pb \(\orbr{\begin{cases}y=\frac{11+5}{2}=8\left(tm\right)\\y=\frac{11-5}{2}=3\left(tm\right)\end{cases}}\)

+) \(y=8\Rightarrow x=9\left(tm\right)\)

+) \(y=3\Rightarrow x=7\left(tm\right)\)

Vậy hệ pt có no (x,y) =( 9,8) ; ( 7,3) 

\(\hept{\begin{cases}x^2+y^2+\frac{2xy}{x+y}=1\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\)\(ĐK:x+y>0\)

Từ \(\left(1\right)\Rightarrow x^2+y^2+2xy-1-2xy+\frac{2xy}{x+y}=0\)

\(\Leftrightarrow\left(x+y\right)^2-1-2xy\left(1-\frac{1}{x+y}\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1\right)-2xy\left(\frac{x+y-1}{x+y}\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1-\frac{2xy}{x+y}\right)=0\)

\(\Leftrightarrow\left(x+y-1\right).\frac{x^2+xy+x+xy+y^2+y-2xy}{x+y}=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2+x+y\right)=0\)( vì x+y >0 )

\(\Leftrightarrow\orbr{\begin{cases}x=1-y\\x^2+y^2+x+y=0\end{cases}}\)

TH1: Xét pt:  \(x^2+y^2+x+y=0\)

Ta thấy \(\hept{\begin{cases}x^2\ge0;\forall x\\y^2\ge0;\forall y\\x+y>0\end{cases}\Rightarrow}x^2+y^2+x+y>0\)

\(\Rightarrow\)pt vô nghiệm

TH2: x=1-y thay vào (2) ta được :

\(\left(1-y\right)^2-y=1\)

\(\Leftrightarrow y^2-3y=0\)

\(\Leftrightarrow y\left(y-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=1\\y=3\Rightarrow x=-2\end{cases}}\)

Vậy hệ pt có no (x,y) = (-2;3) ; (1;0)