\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\)

\(\hept{\begin{cases}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{cases}}\)

\(\hept{\begin{cases}4xy-4xy+8x+12x-6y-12-54=0\\3xy-3xy-3x+3y-3y-3+12=0\end{cases}}\)

\(\hept{\begin{cases}20x-6y-66=0\\-3x+9=0\end{cases}}\)

\(\hept{\begin{cases}2\left(10x-3y\right)=66\\-3\left(x-3\right)=0\end{cases}}\)

\(\hept{\begin{cases}10x-3y=33\\x-3=0\end{cases}}\)

\(\hept{\begin{cases}10x-3y=33\\x=3\end{cases}}\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20x-6y=66\\-3x=-9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

b.

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

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

\(\Leftrightarrow x+3=0\Rightarrow x=-3\Rightarrow y=4\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{2x-5}{3}\\x^2-y^2=40\end{matrix}\right.\)

\(\Rightarrow x^2-\left(\frac{2x-5}{3}\right)^2-40=0\)

\(\Leftrightarrow9x^2-\left(4x^2-20x+25\right)-360=0\)

\(\Leftrightarrow5x^2+20x-385=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\Rightarrow y=3\\x=-11\Rightarrow y=-9\end{matrix}\right.\)

d.

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{36-3x}{2}\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)\left(\frac{36-3x}{2}-3\right)=18\)

\(\Leftrightarrow\left(x-2\right)\left(10-x\right)=12\)

\(\Leftrightarrow-x^2+12x-32=0\Rightarrow\left[{}\begin{matrix}x=4\Rightarrow y=12\\x=8\Rightarrow y=6\end{matrix}\right.\)

e.

\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10m-2\\x-2y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=10m\\x-2y=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2m\\y=m-1\end{matrix}\right.\)

\(x^2-2y^2=1\)

\(\Leftrightarrow4m^2-2\left(m-1\right)^2=1\)

\(\Leftrightarrow4m^2-\left(2m^2-4m+2\right)-1=0\)

\(\Leftrightarrow2m^2+4m-3=0\Rightarrow m=\frac{-2\pm\sqrt{10}}{2}\)

\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

Số 12 nhỏ phía sau không phải đâu các bạn nhé !

5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)

Thay từng TH rồi làm nha bạn

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

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

thay nhá

Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)

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

+) Với y = x - 1 thay vào pt (2):

\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))

Anh quy đồng lên đê, chắc cần vài con trâu đó:))

+) Với y = 2x + 3...