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

<=> \(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)( đến đây đơn giản rồi :)) )

Vậy ...

b) đặt a= 1/x và b = 1/y ( x,y khác 0)

ta có:

15a - 7b =9

4a + 9b = 35 

=> a= 2, b = 3

thay vào ta có:

2 = 1/x => x = 1/2

3 = 1/y => y = 1/3

b) ĐKXĐ : x,y khác 0

Đặt 1/x = a ; 1/y = b ( a,b khác 0 )

hpt đã cho trở thành \(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)(tm)

=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=2\\\dfrac{1}{y}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)(tm)

Vậy ...

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

\(\Leftrightarrow\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-6y+2x-7y=-7+15\\-24x+3y-18x+2y=-3+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-51}{73}\\7x=8+13y=8+13\cdot\dfrac{-51}{73}=-\dfrac{79}{73}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-79}{511}\\y=-\dfrac{51}{73}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{-79}{511}\\y=-\dfrac{51}{73}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\x=-\dfrac{79}{511}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\7x=13y+8=-\dfrac{79}{73}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\x=-\dfrac{79}{511}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

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

=>4x+9y=8 và -8x+3y=5

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

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

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