\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{1}{xy}=6\end{matrix}\right.\left(x,y\ne0\right)\)\(\Leftrightarrow\left(I\right)\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{1}{x}\cdot\dfrac{1}{y}=6\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a,\dfrac{1}{y}=b\left(a,b>0\right)\)

Hệ (I) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\ab=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b\left(5-b\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\-\left(b^2-5b\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^2-5b+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(b-3\right)\left(b-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=5-b\\b-3=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=5-b\\b-2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=5-b\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=5-b\\b=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)

Trả lại biến cũ

\(\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=2\\\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=\dfrac{1}{3}\end{matrix}\right.\left(TM\right)\)và ngược lại

Vậy HPT có các cặp nghiệm là \(\left(0,5;\dfrac{1}{3}\right);\left(\dfrac{1}{3};0,5\right)\)

P/S: Bạn kiểm tra kết quả lại giúp mình nhé

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3.0-y=-3\end{matrix}\right.\)

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

=>12/(x+y-1)-15/(2x-y+3)=15/2 và 12/(x+y-1)-4/(2x-y+3)=28/5

=>x+y-1=22/9; 2x-y+3=-110/19

=>x+y=31/9; 2x-y=-167/19

=>x=-914/513; y=2681/513

Có nhầm đề không bạn ?

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

Đặt x+y-1=a; 2x-y+3=b

Theo đề, ta có: 

4/a-5/b=5/2 và 3/a-1/b=7/5

=>a=22/9; b=-110/19

=>x+y=31/9; 2x+y=-110/19-3=-167/19

=>x=-2092/171; y=2681/171

a) Thay m=1 vào hệ phương trình, ta được:

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+10=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=5\end{matrix}\right.\)

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(-8;5)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=m+4\\x+2\cdot\left(m+4\right)=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2m+8=m+1\\y=m+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-m-7\\y=m+4\end{matrix}\right.\)

Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì \(\left\{{}\begin{matrix}-m-7>3\\m+4< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m>10\\m< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< -10\\m< 1\end{matrix}\right.\Leftrightarrow m< -10\)

Vậy: Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì m<-10

bn giải theo cách thế là ra thôi

hpt

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

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

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

The vao roi tinh la xong