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

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

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

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?

X=0 y=0

Đúng là hỏi rất ngu.

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

Với \(x=0\Leftrightarrow0+2y=0\Leftrightarrow y=0\)

Với \(y=0\Leftrightarrow x=0\)

Vậy \(\left(x,y\right)=\left(0,0\right)\)

a \(\Leftrightarrow\left\{{}\begin{matrix}6x^2-3xy+x=1-y\left(1\right)\\x^2+y^2=1\left(2\right)\end{matrix}\right.\) Từ  (1) \(\Rightarrow6x^2-3xy+x-1+y=0\)

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

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

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

*Nếu 3x-1=0⇔x=\(\dfrac{1}{3}\) Thay vào (2) ta được:

\(\dfrac{1}{9}+y^2=1\Leftrightarrow y^2=\dfrac{8}{9}\Leftrightarrow y=\dfrac{\pm2\sqrt{2}}{3}\)

*Nếu 2x+y=-1\(\Leftrightarrow y=-1-2x\) Thay vào (2) ta được :

\(\Rightarrow x^2+\left(-2x-1\right)^2=1\Leftrightarrow x^2+4x^2+4x+1=1\Leftrightarrow5x^2+4x=0\Leftrightarrow x\left(5x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-4}{5}\end{matrix}\right.\)

.Nếu x=0⇒y=0

.Nếu x=\(\dfrac{-4}{5}\) \(\Rightarrow y=-1+\dfrac{4}{5}=-\dfrac{1}{5}\) Vậy...

Câu b)

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

⇔\(\left\{{}\begin{matrix}2x\left(x-1\right)+y\left(x-1\right)\\x^2-3xy+4=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\left(x-1\right)\left(2x+y\right)=0\\x^2-3xy+4=0\left(2\right)\end{matrix}\right.\)

Để (x-1)(2x+y) = 0 thì: \(\left[{}\begin{matrix}x-1=0\\2x+y=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=1\\2x+y=0\end{matrix}\right.\)

Thay x=1 vào PT (2) ta có:

(2) ⇔1²-3.1.y+4=0

⇔1-3y +4=0

⇔-3y+5=0

⇔y=\(\dfrac{5}{3}\)

Vậy HPT có nghiệm (x:y) = (1;\(\dfrac{5}{3}\))

b\(\left\{{}\begin{matrix}2x^2-2x+xy-y=0\left(1\right)\\x^2-3xy+4=0\left(2\right)\end{matrix}\right.\)  

Từ (1) \(\Rightarrow2x\left(x-1\right)+y\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+y=0\end{matrix}\right.\)

*Nếu x-1=0⇔x=1 Thay vào (2) ta được: \(1-3y+4=0\Leftrightarrow3y=5\Leftrightarrow y=\dfrac{5}{3}\)

*Nếu 2x+y=0\(\Leftrightarrow y=-2x\) Thay vào (2) ta được:

\(\Rightarrow x^2+6x^2+4=0\Leftrightarrow7x^2=-4\) Vô lí ⇒ Trường hợp này ko có x,y (L)

Vậy...

a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

vậy  hệ pt có ndn \(\left\{2;0\right\}\)

b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)

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

vậy hệ pt có ndn \(\left\{2;1\right\}\)

d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)

vậy hệ pt có ndn\(\left\{28;21\right\}\)

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)

a.

Thay số 12 từ pt trên xuống dưới:

\(x^3+2xy^2+y\left(x^2+8y^2\right)=0\)

\(\Leftrightarrow x^3+x^2y+2xy^2+8y^3=0\)

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

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

Thế vào pt đầu:

\(\left(-2y\right)^2+8y^2=12\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)

b.

Thế số 1 từ pt trên xuống dưới:

\(x^7+y^7=\left(x^4+y^4\right)\left(x^3+y^3\right)\)

\(\Leftrightarrow x^4y^3+x^3y^4=0\)

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

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

Thế vào pt đầu: \(\Rightarrow\left[{}\begin{matrix}y^3=1\\x^3=1\\x^3-x^3=1\left(vô-nghiệm\right)\end{matrix}\right.\)

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html

b.

Với \(xy=0\) không là nghiệm

Với \(xy\ne0\)

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)

\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)

\(\Leftrightarrow5-x^2=5x-2x^2\)

\(\Leftrightarrow...\)

c.

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

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

\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=-\dfrac{5x+4}{4}\end{matrix}\right.\)

Thế vào 1 trong 2 pt ban đầu...

X = 96

Y= 80

Z= 48

\(\left\{{}\begin{matrix}x+y+z=224\\-5x+3y+5z=0\\x-2z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y+3z=672\left(1\right)\\-5x+3y+5z=0\left(2\right)\\x-2z=0\left(3\right)\end{matrix}\right.\)

\(\left(1\right)-\left(2\right)\Leftrightarrow8x-2z=672\)

\(\Leftrightarrow4x-z=336\left(4\right)\)

\(\left(3\right);\left(4\right)\Leftrightarrow\left\{{}\begin{matrix}x-2z=0\\4x-z=336\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-8z=0\\4x-z=336\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7z=336\\x-2z=0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x=96\\z=48\end{matrix}\right.\)

\(\Rightarrow y=224-96-48=80\)

Vậy nghiệm hpt đã cho là \(\left\{{}\begin{matrix}x=96\\y=80\\z=48\end{matrix}\right.\)

Lời giải:

Phương hướng giải là bạn sử dụng phương pháp thế, biểu diễn $x$ theo $y$ qua 1 trong 2 PT, sau đó thế vô PT còn lại giải PT 1 ẩn $y$
a) \(\left\{\begin{matrix} x-6y=17\\ 5x+y=23\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=17+6y\\ 5x+y=23\end{matrix}\right.\)

\(\Rightarrow 5(17+6y)+y=23\)

\(\Leftrightarrow 31y=-62\Leftrightarrow y=-2\)

$x=17+6y=17+6(-2)=5$

Vậy $(x,y)=(5,-2)$

Các phần còn lại bạn giải tương tự

b) $(x,y)=(\frac{1}{4}, 0)$

c) $(x,y)=(3, 4)$

d) $(x,y)=(\frac{79}{21}, \frac{44}{21})$