a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)

Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )

b, Làm tương tự a

c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)

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

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

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

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

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

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

Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )

có ái đó giúp mình với mình đang cần gấp

Bài 1:

Lấy PT $(1)$ trừ PT $(2)$ ta có:

\(x^2-y^2=3y-3x\)

\(\Leftrightarrow (x-y)(x+y)+3(x-y)=0\Leftrightarrow (x-y)(x+y+3)=0\)

$\Rightarrow x-y=0$ hoặc $x+y+3=0$

Nếu $x-y=0\Leftrightarrow x=y$. Thay vào PT $(1)$:

\(x^2=3x-2\Leftrightarrow x^2-3x+2=0\Leftrightarrow (x-1)(x-2)=0\)

$\Rightarrow x=1$ hoặc $x=2$

Tương ứng ta thu được $y=1$ hoặc $y=2$

Nếu $x+y+3=0\Leftrightarrow y=-(x+3)$. Thay vào PT $(1)$:

\(x^2=-3(x+3)-2\Leftrightarrow x^2=-3x-11\Leftrightarrow x^2+3x+11=0\)

\(\Leftrightarrow (x+\frac{3}{2})^2=\frac{-35}{4}< 0\) (vô lý)

Vậy..........

Bài 2:

Lấy PT(1) trừ PT(2) ta có:

\(2x-2y+\frac{1}{y}-\frac{1}{x}=\frac{3}{x}-\frac{3}{y}\)

\(\Leftrightarrow 2(x-y)+(\frac{4}{y}-\frac{4}{x})=0\)

\(\Leftrightarrow (x-y)+\frac{2(x-y)}{xy}=0\)

\(\Leftrightarrow (x-y).\frac{2+xy}{xy}=0\Rightarrow \left[\begin{matrix} x=y\\ xy=-2\end{matrix}\right.\)

Nếu $x=y$. Thay vào PT (1) có:

\(2x+\frac{1}{x}=\frac{3}{x}\Leftrightarrow 2x-\frac{2}{x}=0\Leftrightarrow x^2-1=0\)

\(\Rightarrow x^2=1\Rightarrow x=\pm 1\Rightarrow y=\pm 1\) (tương ứng)

Nếu $xy=-2\Rightarrow \frac{1}{y}=\frac{-x}{2}$

Thay vào PT(1): $2x-\frac{x}{2}=\frac{3}{x}$

$\Leftrightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$

$\Rightarrow y=\mp \sqrt{2}$

Vậy........

Bài 3: ĐK: $x,y\neq 0$

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

Lấy PT(1) trừ PT(2) thu được:

\(3xy(x-y)=-(x-y)(x+y)\)

\(\Leftrightarrow 3xy(x-y)+(x-y)(x+y)=0\)

\(\Leftrightarrow (x-y)(3xy+x+y)=0\) \(\Rightarrow \left[\begin{matrix} x=y\\ 3xy=-(x+y)\end{matrix}\right.\)

Nếu $x=y$. Thay vào $(1)$:

\(3x^3=x^2+2\Leftrightarrow 3x^3-x^2-2=0\)

\(\Leftrightarrow (x-1)(3x^2+2x+2)=0\)

Dễ thấy $3x^2+2x+2>0$ nên $x-1=0\Rightarrow x=1\Rightarrow y=1$

Nếu $3xy=-(x+y)$. Lấy $(1)+(2)$ có:

$3xy(x+y)=x^2+y^2+4$

$\Leftrightarrow x^2+y^2+4=-(x+y)^2\leq 0$ (vô lý)

Vậy.......

a) Xem lại đề

b) \(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5.\frac{33-5y}{2}-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}165-25y-6y=10\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31y=155\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=\frac{33-5.5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=4\end{matrix}\right.\)

c)\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{x}{2}-\frac{13-5x}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{3x-26+10x}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5.2\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}x+2y=\frac{7}{6}\\4x+6y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7}{6}-2y\\4\left(\frac{7}{6}-2y\right)+6y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7-12y}{6}\\28-48y+36y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7-12y}{6}\\12y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7-12y}{6}\\y=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7-12.\frac{1}{3}}{6}\\y=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

1/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)

\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)

2/ ĐKXĐ:...

Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)

3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)

4/ Bạn tự giải