a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)

$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$

`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$

`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$

Vậy HPT vô nghiệm

$\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|+2|y-1|=9\\2x+2|y-1|=-2\\\end{cases}$

`<=>` $\begin{cases}|x-2|-2x=11\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|=2x+11\\x+|y-1|=-1\\\end{cases}$

Đến đây dễ rồi bạn tự giải :D

a, Đặt \(\hept{\begin{cases}\frac{1}{x}=u\\\frac{1}{y}=v\end{cases}}\left(u;v\ne0\right)\)

\(\Leftrightarrow\hept{\begin{cases}u+v=\frac{5}{6}\\\frac{1}{6}u+\frac{1}{5}v=\frac{3}{20}\end{cases}}\Leftrightarrow\hept{\begin{cases}u=\frac{5}{6}-v\left(1\right)\\\frac{1}{6}u+\frac{1}{5}v=\frac{3}{20}\left(2\right)\end{cases}}\)

Thay (1) vào (2) ta được : \(\frac{1}{6}\left(\frac{5}{6}-v\right)+\frac{1}{5}v=\frac{3}{20}\)

\(\Leftrightarrow\frac{5}{36}-\frac{v}{6}+\frac{v}{5}=\frac{3}{20}\)

\(\Leftrightarrow\frac{-v}{6}+\frac{v}{5}=\frac{3}{20}-\frac{5}{36}\Leftrightarrow\frac{v}{30}=\frac{1}{90}\Leftrightarrow v=\frac{1}{3}\)(*)

hay \(v=\frac{1}{3}=\frac{1}{y}\Rightarrow y=3\)

Thay (*) vào (1) ta được : \(u=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}\)hay \(u=\frac{1}{2}=\frac{1}{x}\Rightarrow x=2\)

Vậy x = 2 ; y = 3 

b, \(\hept{\begin{cases}4\left(x+y\right)=5\left(x-y\right)\\\frac{40}{x+y}+\frac{40}{x-y}=9\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{x-y}=\frac{5}{x+y}\left(1\right)\\\frac{40}{x+y}+\frac{40}{x-y}=9\left(2\right)\end{cases}}\)

Xét phương trình 1 ta có : \(\frac{4}{x-y}-\frac{5}{x+y}=0\)

\(\Leftrightarrow\frac{4\left(x+y\right)-5\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=0\Leftrightarrow4x+4y-5x+5y=0\)

\(\Leftrightarrow-x+9y=0\Leftrightarrow x=9y\)(*) 

Thay vào 2 ta có : \(\frac{40}{9y+y}+\frac{40}{9y-y}=9\)

\(\Leftrightarrow\frac{4}{y}+\frac{5}{y}=9\Leftrightarrow\frac{9}{y}=9\Leftrightarrow y=1\)

Thay y = 1 vào (*) ta có : \(x=9.1=9\)

Vậy x = 9 ; y = 1

a) $(-3;3),$ $(-3;-1)$.

b) $(16;-7)$.

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)\left(x^2+y^2\right)=26\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)

Trừ vế theo vế \(pt\left(1\right)\) cho \(pt\left(2\right)\) ta được:

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

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

\(\Leftrightarrow x-y=1\)

Khi đó hệ trở thành:

\(\left\{{}\begin{matrix}x^2+y^2=13\\\left(x+y\right)^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\13+2xy=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\2xy=12\end{matrix}\right.\)

Cộng vế theo vế 2 phương trình:

\(\left(x+y\right)^2=25\)

\(\Leftrightarrow x+y=\pm5\)

TH1: \(x+y=5\)

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

TH2: \(x+y=-5\)

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

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

ĐK: \(y\ne0\)

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

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

Đặt \(\dfrac{1}{y}=t\), hệ trở thành:

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

\(\Rightarrow\left(x-t\right)\left(x+t+1\right)=0\)

\(\Leftrightarrow...\)

a: =>2x-4+3+3y=-2 và 3x-6-2-2y=-3

=>2x+3y=-2+4-3=2-3=-1 và 3x-2y=-3+6+2=5

=>x=1; y=-1

b: =>x^2-x+xy-y=x^2+x-xy-y+2xy

=>-x-y=x-y và y^2+y-yx-x=y^2-2y+xy-2x-2xy

=>x=0 và y-x=-2y-2x

=>x=0 và y=0

\(\Leftrightarrow\left\{{}\begin{matrix}xy+6x-3y-18=xy\\xy-2x+2y-4=xy\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)

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

Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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

b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\) 

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

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

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

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4

=>-2x+y=4 và 20x+3y=2

=>x=-5/13; y=42/13

b: =>4x+2|y|=8 và 4x-3y=1

=>2|y|-3y=7 và 4x-3y=1

TH1: y>=0

=>2y-3y=7 và 4x-3y=1

=>-y=7 và 4x-3y=1

=>y=-7(loại)

TH2: y<0

=>-2y-3y=7 và 4x-3y=1

=>y=-7/5; 4x=1+3y=1-21/5=-16/5

=>x=-4/5; y=-7/5