Lời giải:
Ta có:

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

\(\Rightarrow \left\{\begin{matrix} 6x+6y=5xy\\ 20y-15x=5xy\end{matrix}\right.\)

Lấy PT(1) - PT(2):

\(6x+6y-(20y-15x)=0\)

\(\Leftrightarrow 21x=14y\Leftrightarrow 3x=2y\Rightarrow y=1,5x\)

Thay vào PT ban đầu:

\(6x+6.1,5x=5x.1,5x\)

\(\Leftrightarrow 15x=7,5x^2\Leftrightarrow x(7,5x-15)=0\)

Vì $x\neq 0$ nên \(7,5x-15=0\Leftrightarrow x=2\Rightarrow y=1,5.2=3\)

Vậy $(x,y)=(2,3)$

\(\left\{{}\begin{matrix}6x+6y=5xy(1)\\\dfrac{4}{x}-\dfrac{3}{y}=1\end{matrix}\right.\)
Chia 2 vế cho xy thì (1)(vì `x,y ne 0`)
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{4}{x}-\dfrac{3}{y}=1\\\end{cases}$
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{8}{x}-\dfrac{6}{y}=2\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}x=2\\y=3\\\end{cases}$
Vậy HPT có nghiệm (x,y)=(2,3)

\(\left\{{}\begin{matrix}3x-3y=5\\5x+2y=23\end{matrix}\right.< =>\left\{{}\begin{matrix}6x-6y=10\\15x+6y=69\end{matrix}\right.< =>\left\{{}\begin{matrix}21x=79\\3x-3y=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=\dfrac{79}{21}\\y=\dfrac{44}{21}\end{matrix}\right.\)

vậy hệ pt có nghiệm (x,y)=(\(\dfrac{79}{21};\dfrac{44}{21}\))

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

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

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

(Các câu khác tương tự nhé.)

\(\left\{{}\begin{matrix}\dfrac{2y+1}{18}=\dfrac{4y+1}{24}\\\dfrac{4y+1}{24}=\dfrac{6y+1}{6x}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}12y+3-8y-4=0\\\dfrac{4y+1}{4}=\dfrac{6y+1}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\\dfrac{2}{4}=\dfrac{\dfrac{3}{2}+1}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\\dfrac{5}{2}:x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(5;\dfrac{1}{4}\right)\)

a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)

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

c: ĐKXĐ: y>2

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

=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)

Bài 2:

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

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1

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

ĐK: `x ne 2; y ne -1`

Đặt `{a=(1/(x-2)),(b=1/(y+1)):}`

Có: `{(2a+b=3),(4a-3b=1):}`

`<=>{(4a+2b=6),(4a-3b=1):}`

`<=>{(2a+b=3),(5b=5):}`

`<=>{(2a+1=3),(b=1):}`

`<=>{(a=1),(b=1):}`

``

`=>{(1/(x-2)=1),(1/(y+1)=1):}`

`<=>{(x-2=1),(y+1=1):}`

`<=>{(x=3),(y=0):}` (TM)

``

Vậy `(x;y)=(3;0)`.