\(PT\left(2\right)\Leftrightarrow x=y-1\\ PT\left(1\right)\Leftrightarrow2\left(y-1\right)^2+y\left(1-y\right)+3y^2=7\left(y-1\right)+12y-1\\ \Leftrightarrow2y^2-11y+5=0\\ \Leftrightarrow\left[{}\begin{matrix}y=5\Leftrightarrow x=4\\y=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

\(\left\{{}\begin{matrix}2x^2-xy+3y^2-7x-12y+1=0\left(1\right)\\x-y+1=0\left(2\right)\end{matrix}\right.\)

Từ (2) SUY RA : \(x=y-1\)

Thay x = y - 1 vào (1) được :

\(2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\Leftrightarrow2y^2-4y+2-y^2+y+3y^2-7y+7-12y+1=0\Leftrightarrow4y^2-22y+10=0\) \(\Leftrightarrow\) \(\left(y-5\right)\left(2y-1\right)=0\Leftrightarrow\left[{}\begin{matrix}y=5\\y=\dfrac{1}{2}\end{matrix}\right.\)

Suy ra x = 4 hoặc x = -1/2

Vậy nghiệm của hệ pt là (4;5) (-1/2;1/2)

\(x^3-7x^2y+16xy^2-12y^3=0\)

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

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

Thế xuống pt dưới giải đơn giản

\(\begin{cases} x - 4y + 3\sqrt{y} = \sqrt{2x + y} & (1) \\ \sqrt{8y - 1} + x^2 - 12y + 1 = 0 & (2) \end{cases}\)

ĐKXĐ: y>=1/8; x>=-y/2

Đặt \(a=\sqrt{2x+y}\ge0;b=\sqrt{y}\ge\frac{1}{2\sqrt2}>0\)

=>\(a^2=2x+y\)

=>\(x=\frac{a^2 - b^2}{2}\)

(1): \(x - 4y + 3\sqrt{y} = \sqrt{2x + y}\)

=>\(\frac{a^2 - b^2}{2} - 4b^2 + 3b = a\)

=>\(a^2-b^2-8b^2+6b-2a=0\)

=>\(a^2-2a+1-9b^2+6b-1=0\)

=>\(\left(a-1\right)^2-\left(3b-1\right)^2=0\)

=>(a-1-3b+1)(a-1+3b-1)=0

=>(a-3b)(a+3b-2)=0

TH1: a-3b=0

=>a=3b

=>2x+y=9y

=>2x=8y

=>x=4y

Thay x=4y vào (2), ta được:

\(\sqrt{8y - 1} + (4y)^2 - 12y + 1 = 0\)

=>\(\sqrt{8y - 1}+16y^2-12y+1=0\)

Đặt \(t=\sqrt{8y - 1}\ge0\)

=>\(8y=t^2+1\)

=>\(y=\frac{t^2 + 1}{8}\)

\(\sqrt{8y - 1} + 16y^2 - 12y + 1 = 0\)

=>\(t+16\cdot\left(\frac{t^2 + 1}{8}\right)^2-12\cdot\left(\frac{t^2 + 1}{8}\right)+1=0\)

=>\(t+\frac{(t^2 + 1)^2}{4}-\frac{3(t^2 + 1)}{2}+1=0\)

=>\(4t+(t^4+2t^2+1)-6(t^2+1)+4=0\)

=>\(t^4-4t^2+4t-1=0\)

=>\((t^4-1)-4t(t-1)=0\)

=>\((t-1)(t+1)(t^2+1)-4t(t-1)=0\)

=>\((t-1)\left[(t+1)(t^2+1)-4t\right]=0\)

=>\((t-1)(t^3+t^2-3t+1)=0\)

=>\((t-1)^2(t^2+2t-1)=0\)

TH1: t-1=0

=>t=1

=>8y-1=1

=>8y=2

=>y=1/4(nhận)

=>x=4y=1(nhận)

TH2: \(t^2+2t-1=0\)

=>\(t^2+2t+1=2\)

=>\(\left(t+1\right)^2=2\)

=>\(t+1=\sqrt2\) (Do t>0)

=>\(t=\sqrt2-1\)

=>\(8y-1=3-2\sqrt{2}\)

=>\(8y=4-2\sqrt{2}\)

=>\(y=\frac{2 - \sqrt{2}}{4}\) (nhận)

=>\(x=4y=2-\sqrt2\) (nhận)

TH2:a+3b-2=0

=>\(\sqrt{2x+y}+3\sqrt{y}=2\) (3)

y>=1/8

=>\(3\sqrt{y}\ge3\cdot\sqrt{\frac18}=\frac{3}{2\sqrt2}\) ≃1,0606

(2)=>\(x^2 = 12y - 1 - \sqrt{8y - 1}\)

Với y>=1/8, ta xét hàm số \(f\left(y\right)=12y-1-\sqrt{8y - 1}\)

=>f(y)>=0

Khi kết hợp cùng (3), ta sẽ thấy các nghiệm giống hệt với TH1

Vậy: \((x; y) \in \left\{ \left(1; \frac{1}{4}\right), \left(2 - \sqrt{2}; \frac{2 - \sqrt{2}}{4}\right) \right\}\)

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:

\(\dfrac{a^2-b^2}{2}-4b^2+3b=a\Leftrightarrow a^2-9b^2+6b=2a\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b\right)-2\left(a-3b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+3b-2\right)=0\)

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

1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

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

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

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

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)

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