Question

cho hệ phương trình {x+2y=1

{5x+2y=1

b){-2x+y=1

{6x-3y=2

c)\(\frac{3}{5x}\)+\(\frac{1}{y}\)=\(\frac{1}{10}\)

{\(\frac{3}{4x}\)+\(\frac{3}{4y}\)=\(\frac{1}{12}\)

d){\(\frac{8}{x-1}\)+\(\frac{15}{y+2}\)=1

{\(\frac{1}{x-1}\)+\(\frac{1}{y+2}\)=12

Answer 1

a)

\(\left\{{}\begin{matrix}x+2y=1\\5x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x=0\\x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\frac{1}{2}\end{matrix}\right.\)

Answer 2

ý b dễ nên mk giải ý c và d thôi nha

\(\left\{{}\begin{matrix}\frac{3}{5x}+\frac{1}{y}=\frac{1}{10}\\\frac{3}{4x}+\frac{3}{4y}=\frac{1}{12}\end{matrix}\right.\) Đặt \(\frac{3}{x}=a:\frac{1}{y}=b\) ta đcc

\(\left\{{}\begin{matrix}\frac{a}{5}+b=\frac{1}{10}\\\frac{a}{4}+\frac{3b}{4}=\frac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+10=1\\3a+9b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{12}\\b=\frac{1}{12}\end{matrix}\right.\)

thay lại ta được

\(\frac{3}{x}=\frac{1}{12}\Rightarrow x=36\)

\(\frac{1}{y}=\frac{1}{12}\Rightarrow y=12\)

Answer 3

d)

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

\(\left\{{}\begin{matrix}8a+15b=1\\a+b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{179}{7}\\b=\frac{-95}{7}\end{matrix}\right.\)

thay lại ta đc

\(\frac{1}{x-1}=\frac{179}{7}\Leftrightarrow179x=186\Rightarrow x=\frac{186}{179}\)

\(\frac{1}{y+2}=\frac{-95}{7}\Leftrightarrow-95y=197\Rightarrow y=\frac{-195}{7}\)

ý d mk ko bt là đúng hay ko đâu