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

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)

\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)

\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)

Xét phương trình đầu ta có:

2x² + 2y² + 4xy + 3x + 3y - 2 = 0

<=> (2x² + 2xy + 4x) + (2xy + 2y² + 4y) + (- x - y - 2) = 0

<=> (x + y + 2)(2x + 2y - 1) = 0

Giờ chỉ cần thế ngược lại phương trình thứ 2 là giải ra nhé. 

bài này khó quá mong bạn giải giùm mình,mình suy nghĩ hoài mà ko được.

y = 0

x = -1

Ta có hệ:\(\hept{\begin{cases}2x+3y=-2\\3x-2y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}6x+9y=-6\\6x-4y=-6\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}6x+9y=-6\\13y=0\end{cases}\Leftrightarrow\hept{\begin{cases}6x+9.0=6\\y=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}\)

Vậy N⁰ (x;y) củ hệ là:(1;0)

Ấy  chết nhầm ở chỗ :6x+9.0=6 đáng lẽ là 6x+9.0=-6 <=> x=-1 bn sử lại giùm mk nha sorry!^^

\(\left\{{}\begin{matrix}2x+3y=-2\\3x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-9y=-6\\6x-4y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=0\\2x+3y=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\2x+3.0=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm duy nhất là \(\left(x;y\right)=\left(-\frac{3}{2};0\right)\)

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

Trừ trên cho dưới:

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

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

TH1: \(x=y\) thay vào pt đầu ta được \(0=12\) (vô nghiệm)

TH2: \(x=3\Rightarrow-3y^2+3x+6=0\Rightarrow\left[{}\begin{matrix}y=-1\\y=2\end{matrix}\right.\)

TH3: \(y=2\Rightarrow2x^2+2x-24=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Vậy pt có 3 cặp nghiệm \(\left(x;y\right)=\left(3;-1\right);\left(3;2\right);\left(-4;2\right)\)

b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)

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

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

a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}3x+5y=3\\5x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+10y=6\\25x+10y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=3\\19x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-3}{19}+5y=3\\x=\dfrac{-1}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{12}{19}\\x=\dfrac{-1}{19}\end{matrix}\right.\)

Vậy hpt có nghiệm (x;y)=(-1/19;12/19)

b, \(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(x-y\right)=5+3x+2y\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\2x-7y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\4x-14y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}23y=-2\\2x-7y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-2}{23}\\x=\dfrac{101}{46}\end{matrix}\right.\)

Vậy hpt có nghiệm (x;y) = ( 101/46;-2/23)

\(\hept{\begin{cases}2x^4+3x^3+45x=27y^2\left(1\right)\\2y^2-x^2+1=\sqrt{3y^4-4x^2+6y^2-2x^2y^2}\left(2\right)\end{cases}}\)

Xét (2) ta có

\(2y^2-x^2+1=\sqrt{3y^4-4x^2+6y^2-2x^2y^2}\)

Bình phương 2 vế rút gọn ta được

\(\Leftrightarrow y^4+x^4-2x^2y^2-2y^2+2x^2+1=0\)

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

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

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

Thế (3) vào (1) ta được

\(2x^4+3x^3+45x=27\left(x^2+1\right)^2\)

\(\Leftrightarrow25x^4-3x^3+54x^2-45x+27=0\)

\(\Leftrightarrow\left(25x^4-\frac{2.5.3}{2.5}x^3+\frac{9}{100}x^2\right)+\left(\frac{5391}{100}x^2-\frac{2\sqrt{5391}.45.10}{10.\sqrt{5391}.2}x+\frac{5625}{599}\right)+\frac{10548}{599}=0\)

\(\Leftrightarrow\left(5x^2-\frac{3}{10}x\right)^2+\left(\frac{\sqrt{5391}}{10}x-\frac{45}{\sqrt{599}}\right)^2+\frac{10548}{599}=0\)

\(\Rightarrow\)PT vô nghiệm

PS: Đề có sai không mà nhìn gớm vậy bạn

\(\hept{\begin{cases}2x^4+3x^3+45x=27y^2\left(1\right)\\2y^2-x^2+1=\sqrt{3y^4-4x^2+6y^2-2x^2y^2}\left(2\right)\end{cases}}\)

ĐK: \(2y^2+1\ge1\)

Phương trình (2) tương đương:

\(\left(2y^2-x^2+1\right)^2=3y^4-4x^2+6y^2-2x^2y^2\)

\(\Leftrightarrow y^4+2x^2-2x^2y^2+x^4+1-2y^2=0\)

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

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

Thế \(x^2+1=y^2\) vào phương trình (1) ta có:

\(2x^4+3x^3+45x=27\left(x^2+1\right)\)

\(\Leftrightarrow2x^4+3x^3-27x^2+45x-27=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)\left(2x^3+6x^2-18x+18\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\Rightarrow y=\frac{\sqrt{13}}{2}\\x=-\sqrt[3]{16}-\sqrt[3]{4}-1\Rightarrow y=\sqrt{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2+1}\end{cases}}\)

Vậy:.....

Bạn alibaba nguyễn làm sai ở bước thay y^2=x^2+1 rồi