đk: \(x\ge0;y\ge-1\)

\(pt\left(1\right)\Leftrightarrow y^2-\left(x^2-5x-1\right)y-\left(x^3-3x^2-4x\right)=0\)

\(\Leftrightarrow\left(y+x+1\right)\left(y-x^2+4x\right)=0\Leftrightarrow\orbr{\begin{cases}y=x^2-4x\\y+x+1=0\end{cases}}\)

Từ pt(2) \(\Leftrightarrow3\sqrt{x}=\sqrt{y+1}+x+1\ge1\Rightarrow x>0\Rightarrow y+x+1>0\)

Vậy ta có \(\left(1\right)\Leftrightarrow y=x^2-4x\)

Thay \(y=x^2-4x\)vào (1) ta có: \(3\sqrt{x}-\sqrt{x^2-4x+1}=x+1\left(3\right)\)

Vì x=0 không là nghiệm của (3) nên \(\left(3\right)\Leftrightarrow\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x+\frac{1}{x}-4}=3\)

Đặt \(t=\sqrt{x}+\frac{1}{\sqrt{x}}\left(t\ge2\right)\Rightarrow x+\frac{1}{x}=t^2-2\). PT trở thành:

\(t+\sqrt{t^2-6}=3\Leftrightarrow\sqrt{t^2-6}=3-t\Leftrightarrow\hept{\begin{cases}t\le3\\t^2-6=\left(3-t\right)^2\end{cases}}\Leftrightarrow t=\frac{5}{2}\)

\(\Leftrightarrow x+\frac{1}{x}=\frac{25}{4}-2\Leftrightarrow x^2-\frac{17}{4}x+1=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{4}\end{cases}}\)

Từ đó suy ra hệ pt có 2 nghiệm: \(\left(4;0\right);\left(\frac{1}{4};\frac{-15}{16}\right)\)

\(\left\{{}\begin{matrix}x^2+xy+y^2=1\\x-y-xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2+3xy=1\\x-y-xy=3\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+3v=1\\u-v=3\end{matrix}\right.\)

\(\Rightarrow u^2+3\left(u-3\right)=1\)

\(\Leftrightarrow u^2+3u-10=0\Rightarrow\left[{}\begin{matrix}u=2\Rightarrow v=-1\\u=-5\Rightarrow v=-8\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}u=2\\v=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=2\\xy=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\xy=-1\end{matrix}\right.\)

\(\Rightarrow x\left(x-2\right)=-1\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\Rightarrow y=-1\)

TH2: \(\left\{{}\begin{matrix}u=-5\\v=-8\end{matrix}\right.\) \(\Rightarrow...\) bạn tự làm tương tự

a ) \(y\left(x-1\right)=x^2+2\)

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

\(\Leftrightarrow x^2-1-y\left(x-1\right)+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)=-3\)

\(\Leftrightarrow\left(x-1\right)\left(x+1-y\right)=-3\)

...

b ) \(3xy-5x-2y=3\)

\(\Leftrightarrow9xy-15x-6y=9\)

\(\Leftrightarrow9xy-15x-6y+10=19\)

\(\Leftrightarrow3y\left(3x-2\right)-5\left(3x-2\right)=19\)

\(\Leftrightarrow\left(3y-5\right)\left(3x-2\right)=19\)

...

c ) \(x^2-10xy-11y^2=13\)

\(\Leftrightarrow x^2-11xy+xy-11y^2=13\)

\(\Leftrightarrow x\left(x-11y\right)+y\left(x-11y\right)=13\)

\(\Leftrightarrow\left(x+y\right)\left(x-11y\right)=13\)

...

d ) \(xy-2=2x+3y\)

\(\Leftrightarrow xy-2-2x-3y=0\)

\(\Leftrightarrow y\left(x-3\right)-2\left(x-3\right)-8=0\)

\(\Leftrightarrow\left(y-2\right)\left(x-3\right)=8\)

...

e ) \(5xy+x+2y=7\)

\(\Leftrightarrow5xy+x+2y-7=0\)

\(\Leftrightarrow5x\left(y+\dfrac{1}{5}\right)+2\left(y+\dfrac{1}{5}\right)-\dfrac{37}{5}=0\)

\(\Leftrightarrow\left(5x+2\right)\left(y+\dfrac{1}{5}\right)=\dfrac{37}{5}\)

\(\Leftrightarrow\left(5x+2\right)\left(5y+1\right)=37\)

...

P/s : Vì bài dài nên việc tìm x , y ( lập bảng ) bạn tự làm nhé

Thanks

\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}=\dfrac{x+3}{2019}+\dfrac{x+4}{2018}\)

=>\(\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

=> (x+2022)(\(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\))=0

=>x+2022=0

=> x=-2022

\(x+\sqrt{9-x^2}-x\sqrt{9-x^2}=3\left(-3\le x\le3\right)\)

\(\Leftrightarrow\sqrt{9-x^2}-x\sqrt{9-x^2}=3-x\\ \Leftrightarrow9-x^2+x^2\left(9-x^2\right)-2x\sqrt{\left(9-x^2\right)^2}=9-6x+x^2\\ \Leftrightarrow9+8x^2-x^4-2x\left(9-x^2\right)=x^2-6x+9\\ \Leftrightarrow-x^4+2x^3+7x^2-12x=0\\ \Leftrightarrow-x\left(x^3-2x^2-7x+12\right)=0\Leftrightarrow-x\left(x^3-3x^2+x^2-3x-4x+12\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2+x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=3\left(N\right)\\x^2+x-4=0\left(1\right)\end{matrix}\right.\)

 \(\Delta\left(1\right)=1-4\left(-4\right)=17>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-1+\sqrt{17}}{2}\left(N\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;3;\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)

Tick ✔