1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

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

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x^3+2x^2y+xy+y^2+2x^2+x+2y=7\\4x^2+x+3y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(4x+1\right)+3y=7\\2x^3+xy+2x^2y+y^2+2x^2+x+2y-4x^2-x-3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(4x+1\right)+3y=7\\2x^3+xy+2x^2y+y^2-2x^2-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+x+3y=7\\x\left(2x^2+y\right)+y\left(2x^2+y\right)-\left(2x^2+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+x+3y=7\\\left(2x^2+y\right)\left(x+y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+x+3y=7\\\left(2x^2+y\right)\left(x+y-1\right)=0\end{matrix}\right.\)

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

Xét TH1:\(2x^2=-y\) (vô lý) =.> Loại

Xét TH2: y=1-x

Thay \(y=1-x\) vào (1) ta được :

(1)\(\Leftrightarrow4x^2+x+3\left(1-x\right)=7\)

\(\Leftrightarrow4x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{17}}{4}\\x_2=\dfrac{1-\sqrt{17}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x1=\dfrac{1+\sqrt{17}}{4}\\y1=\dfrac{3-\sqrt{17}}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x2=\dfrac{1-\sqrt{17}}{4}\\y2=\dfrac{3+\sqrt{17}}{4}\end{matrix}\right.\end{matrix}\right.\)

KL: phương trình (I) có 2 nghiệm là (x;y)=........

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 ...

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

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

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