Bài 23:
a: \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2xy+5x-6y-5=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x-6y+2x-7y=-7+5\\-24x+3y-18x+2y=-3+6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-13y=-2\\-42x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42x-78y=-12\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}42x-78y-42x+5y=-12+3\\7x-13y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-73y=-9\\7x=13y-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{9}{73}\\x=\dfrac{13y-2}{7}=\dfrac{13\cdot\dfrac{9}{73}-2}{7}=-\dfrac{29}{511}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x-y=x-y\\y-x=-2y-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x=0\\3y=-x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Bài 24:
a: ĐKXĐ: \(x\ne0;y\ne0\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}a+b=\dfrac{4}{5}\\a-b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+a-b=\dfrac{4}{5}+\dfrac{1}{5}\\a-b=\dfrac{1}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2a=1\\b=a-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: \(x\ne0;y\ne0\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}60a-28b=36\\60a+135b=525\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}60a-28b-60a-135b=36-525\\15a-7b=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-163b=-489\\15a=7b+9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\a=\dfrac{7b+9}{15}=\dfrac{7\cdot3+9}{15}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: \(x\ne\pm y\)
Đặt \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}a+b=\dfrac{5}{8}\\a-b=-\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+a-b=\dfrac{5}{8}-\dfrac{3}{8}\\a+b=\dfrac{5}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2a=\dfrac{2}{8}\\b=\dfrac{5}{8}-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{8}\\b=\dfrac{5}{8}-\dfrac{1}{8}=\dfrac{4}{8}=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=8\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x+y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\left(nhận\right)\)
d: ĐKXĐ: \(y\ne-3x;y\ne\dfrac{2}{3}x\)
Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}4a+5b=-2\\-5a+3b=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20a+25b=-10\\-20a+12b=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20a+25b-20a+12b=-10+84\\4a+5b=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}37b=74\\4a=-5b-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=\dfrac{-5b-2}{4}=\dfrac{-5\cdot2-2}{4}=-\dfrac{12}{4}=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\9x+3y=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3y+9x+3y=-\dfrac{1}{3}+\dfrac{3}{2}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=-\dfrac{2}{6}+\dfrac{9}{6}=\dfrac{7}{6}\\y=\dfrac{1}{2}-3x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{1}{2}-3\cdot\dfrac{7}{66}=\dfrac{1}{2}-\dfrac{7}{22}=\dfrac{4}{22}=\dfrac{2}{11}\end{matrix}\right.\)(nhận)
e: ĐKXĐ: \(x\ne y-2;x\ne-y+1\)
Đặt \(\dfrac{1}{x-y+2}=a;\dfrac{1}{x+y-1}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}7a-5b=4,5\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14a-10b=9\\15a+10b=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}14a-10b+15a+10b=9+20\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}29a=29\\2b=4-3a\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=1\\b=\dfrac{4-3a}{2}=\dfrac{4-3}{2}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-y+2}=1\\\dfrac{1}{x+y-1}=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+2=1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=3-x=3-1=2\end{matrix}\right.\)(nhận)
