Question

[nguyen thi ngoc anh]

GIÚP MK VỚI 

GIẢI HỆ PT 

A;

\(\frac{5}{x-1}+\frac{1}{y-1}=10\)

\(\frac{1}{x-1}-\frac{3}{y-1}=18\)

B;

\(\frac{4}{x+2y}-\frac{1}{x-2y}=1\)

\(\frac{20}{x+2y}+\frac{3}{x-2y}=1\)

C;

\(\frac{12}{x-3}-\frac{5}{y+2}=63\)

\(\frac{8}{x-3}+\frac{15}{y+2}=-13\)

D;

\(\frac{5}{x+y-3}-\frac{2}{x-y+1}=8\)

\(\frac{3}{x+y-3}+\frac{1}{x-y+1}=1,5\)

Answer 1

[Hoàng Phúc]

a)Đặt \(\frac{1}{x-1}=t;\frac{1}{y-1}=m\)

Ta có: \(\frac{5}{x-1}+\frac{1}{y-1}=10=5.\frac{1}{x-1}+\frac{1}{y-1}=10=5t+m=10\)

\(\frac{1}{x-1}+\frac{3}{y-1}=t+3.\frac{1}{y-1}=t+3m=18\)

Từ đây ta có HPT \(\hept{\begin{cases}5t+m=10\left(1\right)\\t+3m=18\left(2\right)\end{cases}}\)

\(5t+m=10\Rightarrow5t=10-m\Rightarrow t=\frac{10-m}{5}\),thay vào (2) ta có:

\(\frac{10-m}{5}+3m=18\Rightarrow\frac{10-m+15m}{5}=18\Rightarrow\frac{10+14m}{5}=18\)

=>10+14m=18.5=90=>14m=90-10=>14m=80=>m=\(\frac{40}{7}\)

Thay m=40/7 vào (1)=>t=6/7

Vì \(\frac{1}{x-1}=t\Rightarrow\frac{1}{x-1}=\frac{6}{7}\Rightarrow\left(x-1\right).6=7\Rightarrow6x-6=7\Rightarrow x=\frac{13}{6}\)

Vì \(\frac{1}{y-1}=m\Rightarrow\frac{1}{y-1}=\frac{40}{7}\Rightarrow\left(y-1\right).40=7\Rightarrow40y-40=7\Rightarrow y=\frac{47}{40}\)

Vậy x=13/6;y=47/40 thì thỏa mãn HPT

mk hết hè lên lp 8 nên cũng không chắc 100% nhé

Answer 2

[Nguyễn Quang Trung]

b/ Đặt \(\frac{1}{x+2y}=a\) ; \(\frac{1}{x-2y}=b\) , ta có hệ phương trình: \(\hept{\begin{cases}4a-b=1\\20a+3b=1\end{cases}\Rightarrow\hept{\begin{cases}b=4a-1\\20a+3\left(4a-1\right)=1\end{cases}\Rightarrow}\hept{\begin{cases}b=4a-1\\20a+12a-3=1\end{cases}}\Rightarrow\hept{\begin{cases}b=4a-1\\a=\frac{1}{8}\end{cases}\Rightarrow}\hept{\begin{cases}b=-\frac{1}{2}\\a=\frac{1}{8}\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x-2y}=-\frac{1}{2}\\\frac{1}{x+2y}=\frac{1}{8}\end{cases}\Rightarrow\hept{\begin{cases}x-2y=-2\\x+2y=8\end{cases}\Rightarrow}\hept{\begin{cases}x=-2+2y\\-2+2y+2y=8\end{cases}\Rightarrow}\hept{\begin{cases}x=-2+2y\\y=\frac{5}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=\frac{5}{2}\end{cases}}}\)

           Vậy x = 3 , y = 5/2 

c/ Đặt \(\frac{1}{x-3}=a\) ; \(\frac{1}{y+2}=b\) , ta có hệ phương trình: 

     \(\hept{\begin{cases}12a-5b=63\\8a+15b=-13\end{cases}\Rightarrow\hept{\begin{cases}b=\frac{12a-63}{5}\\8a+15\left(\frac{12a-63}{5}\right)=-13\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}b=\frac{12a-63}{5}\\8a+\frac{180a-945}{5}=-13\end{cases}}\)

       \(\Rightarrow\hept{\begin{cases}b=\frac{12a-63}{5}\\a=4\end{cases}\Rightarrow\hept{\begin{cases}b=-3\\a=4\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}\frac{1}{y+2}=-3\\\frac{1}{x-3}=4\end{cases}\Rightarrow\hept{\begin{cases}-3y-6=1\\4x-12=1\end{cases}}\Rightarrow\hept{\begin{cases}y=-\frac{7}{3}\\x=\frac{13}{4}\end{cases}}}\)

             Vậy x = 13/4 , y = -7/3

d/ Đặt \(\frac{1}{x+y-3}=a\) ; \(\frac{1}{x-y+1}=b\) , ta có hệ phương trình:

         \(\hept{\begin{cases}5a-2b=8\\3a+b=1,5\end{cases}\Rightarrow\hept{\begin{cases}5a-2\left(\frac{3}{2}-3a\right)=8\\b=\frac{3}{2}-3a\end{cases}\Rightarrow}\hept{\begin{cases}5a-3+6a=8\\b=\frac{3}{2}-3a\end{cases}\Rightarrow}\hept{\begin{cases}a=1\\b=-\frac{3}{2}\end{cases}}}\)

         \(\Rightarrow\hept{\begin{cases}\frac{1}{x+y-3}=1\\\frac{1}{x-y+1}=-\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}x+y-3=0\\-3x+3y-3=2\end{cases}\Rightarrow}\hept{\begin{cases}x+y=3\\-3x+3y=5\end{cases}}}\)

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

                          Vậy x = 2/3 ; y = 7/3