a) \(\left\{{}\begin{matrix}2x^2-5xy-y^2=1\\y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=1\end{matrix}\right.\)

ĐKXĐ:...

\(\Rightarrow y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=2x^2-5xy-y^2\)

Từ giả thiết dễ thấy \(y\ne0\), chia cả 2 vế cho \(y^2\) ta được:

\(\dfrac{\sqrt{xy-2y^2}+\sqrt{4y^2-xy}}{y}=\dfrac{2x^2-5xy-y^2}{y^2}\)

\(\Leftrightarrow\sqrt{\dfrac{xy-2y^2}{y^2}}+\sqrt{\dfrac{4y^2-xy}{y^2}}=2\left(\dfrac{x}{y}\right)^2-\dfrac{5x}{y}-1\)

\(\Leftrightarrow\sqrt{\dfrac{x}{y}-2}+\sqrt{4-\dfrac{x}{y}}=2\left(\dfrac{x}{y}\right)^2-5\dfrac{x}{y}-1\)

Đặt \(\dfrac{x}{y}=t\) \(\left(2\le t\le4\right)\)

\(\Leftrightarrow\sqrt{t-2}+\sqrt{4-t}=2t^2-5t-1\)

\(\Leftrightarrow\sqrt{t-2}-1+\sqrt{4-t}-1=2t^2-5t-3\)

\(\Leftrightarrow\left(t-3\right)\left(2t+1\right)=\dfrac{t-3}{\sqrt{t-2}+1}+\dfrac{3-t}{\sqrt{4-t}+1}\)

\(\Leftrightarrow\left(t-3\right)\left(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}\right)=0\)

Xét \(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}=2t+\dfrac{\sqrt{t-2}}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}>0\forall t\)

\(\Rightarrow t-3=0\)

\(\Leftrightarrow t=3\)

\(\Leftrightarrow\dfrac{x}{y}=3\Leftrightarrow x=3y\)

Thế vào phương trình \(\left(1\right):2\cdot9y^2-5y\cdot3y-y^2-1=0\)

\(\Leftrightarrow2y^2-1=0\)

\(\Leftrightarrow y=\dfrac{1}{\sqrt{2}}\) do \(y>0\)

\(\Leftrightarrow x=\dfrac{3}{\sqrt{2}}\)

Vậy tập nghiệm của phương trình \(\left(x;y\right)=\left(\dfrac{3}{\sqrt{2}};\dfrac{1}{\sqrt{2}}\right)\)

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

Trừ theo vế 2 phương trình ta được:

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

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-2\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-2\left(x+y\right)+4\right)=0\)

Xét phương trình \(x^2+x\left(y-2\right)+y^2-2y+4=0\)

\(\Delta_x=\left(y-2\right)^2-4\left(y^2-2y+4\right)=-3y^2+4y-8< 0\) nên phương trình vô nghiệm.

Do đó \(x=y\)

Thế vào phương trình \(\left(1\right):x^3+1=2x^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy...

a, ĐK: \(x,y\ge0\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=\sqrt{x+3}\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+3}=x+1\)

\(\Leftrightarrow x+3=x^2+2x+1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)

Thay \(x=1\) vào hệ phương trình đã cho ta được \(y=1\)

Vậy pt đã cho có nghiệm \(x=y=1\)

b, \(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(y+\dfrac{1}{2}\right)^2\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-3x=0\end{matrix}\right.\left(1\right)\\\left\{{}\begin{matrix}x+y=-1\\x^2+y^2=-3\end{matrix}\right.\left(vn\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=0\end{matrix}\right.\)

Vậy ...

c, Đặt \(\left\{{}\begin{matrix}x^2+y^2=a\\xy=b\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a^2-b^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a-b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x+y=\pm3\)

TH1: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)