\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-448y^3=-3x+6y\\96=385x^2-16y^2\end{matrix}\right.\)

\(\Rightarrow96\left(x^3-448y^3\right)=\left(-3x+6y\right)\left(385x^2-16y^2\right)\)

\(\Leftrightarrow\left(x-4y\right)\left(417x^2+898xy+3576y^2\right)=0\)

\(\Leftrightarrow x-4y=0\)

\(\Leftrightarrow x=4y\)

Thế vào \(385x^2-16y^2=96\)

\(\Rightarrow...\)

b.

ĐKXĐ: \(x+y\ne0\)

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

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

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

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

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

Thế vào \(x^2+y^2=1\)...

a, Cộng vế theo vế hai phương trình ta được:

\(x^2+y^2+2xy+x+y=2\)

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

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

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

TH1: \(x+y=1\)

\(pt\left(2\right)\Leftrightarrow xy+1=-1\Leftrightarrow xy=-2\)

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

TH2: \(x+y=-2\)

\(pt\left(2\right)\Leftrightarrow xy-2=-1\Leftrightarrow xy=1\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=-2\\xy=1\end{matrix}\right.\Leftrightarrow x=y=-1\)

b, \(\left\{{}\begin{matrix}x^3-y^3=7\left(x-y\right)\\x^2+y^2=x+y+2\end{matrix}\right.\)

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

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

TH1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2=x+y+2\end{matrix}\right.\)

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

\(\Leftrightarrow x=y=\dfrac{1\pm\sqrt{5}}{2}\)

TH2: \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2=x+y+2\end{matrix}\right.\)

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

Đặt \(x+y=u;xy=v\)

Hệ trở thành: \(\left\{{}\begin{matrix}u^2-v=7\\u^2-2v-u=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2-2\left(u^2-7\right)-u=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2+u-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\\left[{}\begin{matrix}u=3\\u=-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}v=2\\u=3\end{matrix}\right.\\\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\end{matrix}\right.\)

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

Với \(\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=9\\x+y=-4\end{matrix}\right.\left(vn\right)\)

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

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

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

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

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

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

\(\Leftrightarrow5y^4-6y^2+1=0\)

Bài 1:

Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{2x-2}\\v=\dfrac{1}{y-1}\end{matrix}\right.\) (ĐK: \(x,y\ne1\))  

Hệ trở thành:

\(\Leftrightarrow\left\{{}\begin{matrix}u-v=2\\3u-2v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3u-3v=6\\3u-2v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-v=5\\u-v=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=-5\\u=2+-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=-5\\u=-3\end{matrix}\right.\)

Trả lại ẩn của hệ pt:

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-1}=-5\\\dfrac{1}{2x-2}=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=-\dfrac{1}{5}\\2x-2=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x=\dfrac{5}{6}\end{matrix}\right.\left(tm\right)\)

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

a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)

=>x=2 và y=-2

- Với \(xy=0\) không phải nghiệm

- Với \(xy\ne0\) hệ tương đương

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

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

\(\Rightarrow x^3+3x=z^3+3z\)

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

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

\(\Leftrightarrow x=z\)

Thế vào \(x^3+2=3z\Rightarrow x^3+2=3x\)

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

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

a, \(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^2+y^2-xy\right)=70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\3x^2y^2-40xy+93=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[{}\begin{matrix}xy=\dfrac{31}{3}\\xy=3\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{31}{3}\end{matrix}\right.\)

Phương trình này vô nghiệm

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)

b, ĐK: \(xy>0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{y}+\dfrac{2y}{x}+4=9\\x-y+xy=3\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

Vậy ...