\(a\orbr{x=\frac{\pm\sqrt{5}-3}{4}}\)

\(b\hept{\begin{cases}x=5\\y=4\end{cases}}\)

2)\(\Leftrightarrow\left(x^3-x^2y\right)+\left(y^3-xy^2\right)=5\)

\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(y-x\right)=5\)

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

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

TH1\(\hept{\begin{cases}x-y=1\\x^2-y^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}\left(N\right)}}\)

TH2\(\hept{\begin{cases}x-y=5\\x^2-y^2=1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)

TH3\(\hept{\begin{cases}x-y=-1\\x^2-y^2=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}\left(N\right)}}\)

TH4\(\hept{\begin{cases}x-y=-5\\x^2-y^2=-1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)

Vậy......

bạn mai anh làm đúng rồi mình xét thiếu trường hợp . nhưng nên phân tích thành (x+y)(x-y)² dễ hơn

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

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.

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

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

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

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

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

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

ĐK: \(x\ge\frac{1}{2}\)

\(\hept{\begin{cases}x\left(2x-2y-1\right)=3\left(y+2\right)\left(1\right)\\3y+6\sqrt{2x-1}=y^2-x+23\left(2\right)\end{cases}}\)

pt (1) <=> \(2x^2-2xy-x-3y-6=0\)

<=> \(2x^2-x\left(2y+1\right)-\left(3y+6\right)=0\)

có \(\Delta=\left(2y+1\right)^2+4\left(3y+6\right)=4y^2+28y+49=\left(2y+7\right)^2\)

=> (1) có hai nghiệm: \(\orbr{\begin{cases}x_1=\frac{\left(2y+1\right)-\left(2y+7\right)}{4}=-\frac{3}{2}\left(loai\right)\\x_2=\frac{\left(2y+1\right)+\left(2y+7\right)}{4}=y+2\end{cases}}\)

+) Với \(x=y+2\) thế vào (2) ta có: 

\(3y+6\sqrt{2\left(y+2\right)-1}=y^2-\left(y+2\right)+23\)

<=> \(6\sqrt{2y+3}=y^2-4y+21\)

ĐK: \(y\ge-\frac{3}{2}\)

\(6\sqrt{2y+3}=y^2-4y+21\)

<=> \(6\sqrt{2y+3}-2y-12=y^2-6y+9\)

<=> \(\frac{2\left(9\left(2y+3\right)-\left(y+6\right)^2\right)}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\frac{-2\left(y-3\right)^2}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\left(y-3\right)^2\left(\frac{-2}{3\sqrt{2y+3}+y+6}-1\right)=0\)

<=> y - 3 = 0 

<=> y = 3 thỏa mãn 

khi đó x = y + 2 = 3 + 2 = 5 thỏa mãn

Kết luận:...

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

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=2\\2a+3b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2.-1\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Vậy nghiệm hpt \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)