Những bài còn lại chỉ cần phân tích ra rồi rút gọn là được nha. Bạn tự làm nha!

Đặt \(\hept{\begin{cases}x+y=a\\x-y=b\end{cases}}\)\(\Rightarrow\)ta có hệ \(\hept{\begin{cases}2a+3b=4\\a+2b=5\end{cases}}\Rightarrow\hept{\begin{cases}a=-7\\b=6\end{cases}}\)Từ đó ta có \(\hept{\begin{cases}x+y=-7\\x-y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{13}{2}\end{cases}}\)PS: Cái đề chỗ 3(x+y) phải thành 3(x-y) chứ

2) Từ hệ ta có \(\hept{\begin{cases}20x-6y=66\\-3x=-9\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\5x-4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{5}{17}\end{matrix}\right.\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{8}{y}=18\\\dfrac{10}{x}+\dfrac{8}{y}=102\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{x}=120\\\dfrac{1}{x}-\dfrac{8}{y}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{120}\\y=-\dfrac{44}{39}\end{matrix}\right.\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{30}{x-1}+\dfrac{3}{y+2}=3\\\dfrac{25}{x-1}+\dfrac{3}{y+2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}=1\\\dfrac{10}{y-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\)

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

d: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{135}{2x-y}+\dfrac{160}{x+3y}=35\\\dfrac{135}{2x-y}-\dfrac{144}{x+3y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=8\\2x-y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+6y=16\\2x-y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=5\end{matrix}\right.\)

\(\hept{\begin{cases}\frac{7}{x-y+2}-\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{cases}}\)

Đặt \(a=\frac{1}{x-y+2};b=\frac{1}{x+y-1}\)ta được hệ phương trình:

\(\hept{\begin{cases}7a-5b=\frac{9}{2}\\3a+2b=4\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=\frac{1}{2}\end{cases}}}\)

Với \(\hept{\begin{cases}a=1\\b=\frac{1}{2}\end{cases}}\), ta được:

\(\hept{\begin{cases}\frac{1}{x-y+2}=1\\\frac{1}{x+y-1}=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y+2=1\\x+y-1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Vậy hệ phương trình có 1 nghiệm là x = 1 và y = 2 

\(a,\hept{\begin{cases}x^2-3y=2\\9y^2-8x=8\end{cases}}\)

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

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

\(9-\left(\frac{x^2-2}{3}\right)^2-8x=8\)

\(\Rightarrow x^4-4x^2+4-8x-8=0\)

\(\Rightarrow x^4-4x^2-8x-4=0\)

\(\Rightarrow\left(x^2-2x-2\right)\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{2+2\sqrt{3}}{3}\\y=\frac{2-2\sqrt{3}}{3}\end{cases}}\)

Vậy ................................

\(\hept{\begin{cases}\frac{3}{5}x-\frac{2}{5}y+\frac{5}{3}x-y-x=1\\\frac{2}{3}x-y+2x-\frac{3}{2}y-y=1\end{cases}}\)<=>\(\hept{\begin{cases}\frac{19}{15}x-\frac{7}{5}y=1\\\frac{8}{3}x-\frac{7}{2}y=1\end{cases}}\)<=>x=3;y=2

\(\left\{{}\begin{matrix}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{2y-5x}{3}+5+2x=\frac{y+27}{4}\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{2y+x+15}{3}=\frac{y+27}{4}\\\frac{x+3y+1}{3}=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y+4x+60=3y+81\\7x+21y+7=18y-15x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+5y=21\\66x+9y=-21\end{matrix}\right.\Leftrightarrow70x+14y=0\Leftrightarrow5x+y=0\Leftrightarrow20x+4y=0;4x+5y=21\Leftrightarrow20x+25y=105\Leftrightarrow\left(20x+25y\right)-\left(20x+4y\right)=105\Leftrightarrow21y=105\Leftrightarrow y=5.\text{Thay vào ta được:}4x+25=21\Leftrightarrow4x=-4\Leftrightarrow x=-1\)

\(\text{Thử lại ta thấy thỏa mãn: Vậy: x=-1;y=5}\)

\(\left\{{}\begin{matrix}\frac{2}{3}y-\frac{5}{3}x-\frac{1}{4}y+2x=\frac{27}{4}-5\\\frac{1}{3}x+\frac{5}{7}x+y-\frac{6}{7}y=-\frac{1}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{3}x+\frac{5}{12}y=\frac{7}{4}\\\frac{22}{21}x+\frac{1}{7}y=-\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\)

Ôi trời nhiều thía ? làm từng câu một ha !

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

\(\Leftrightarrow\hept{\begin{cases}xy-2x+5y-10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+3y=8\\3x-y=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y-3x+9y=16+24\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\8y=40\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=7\\y=5\end{cases}}\)

b, ĐKXĐ \(x\ne\pm y\)

Đặt \(\frac{1}{x+y}=a\)  và  \(\frac{1}{x-y}=b\)(a và b khác 0)

Ta có hệ \(\hept{\begin{cases}a-2b=2\\5a-4b=3\end{cases}}\)

          \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b=3\end{cases}}\)

       \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b-2a+4b=3-4\end{cases}}\)

       \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\3a=-1\end{cases}}\)

      \(\Leftrightarrow\hept{\begin{cases}a=-\frac{1}{3}\\b=-\frac{7}{6}\end{cases}}\)

    \(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+y}=-\frac{1}{3}\\\frac{1}{x-y}=-\frac{7}{6}\end{cases}}\)

   \(\Leftrightarrow\hept{\begin{cases}x+y=-3\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}x+y-x+y=-3+\frac{6}{7}\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}2y=-\frac{15}{7}\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}x=-\frac{27}{14}\\y=-\frac{15}{14}\end{cases}}\)

c,\(\hept{\begin{cases}4x^2+y^2=13\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x^2+y^2+2x^2-y^2=13-7\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6x^2=6\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=9\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\pm1\\y=\pm3\end{cases}}\)