Question

[22 - Đỗ Nhật Minh - 6A17]

Answer 1

[Nguyễn Lê Phước Thịnh]

a:

ĐKXĐ: \(x\ne0;y\ne0\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành \(\left\{{}\begin{matrix}a+b=\dfrac{4}{5}\\a-b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=1\\a-b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=a-\dfrac{1}{5}=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\\\dfrac{1}{y}=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\left(nhận\right)\)

b: ĐKXĐ: \(x\ne0;y\ne0\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}60a-28b=36\\60a+135b=140\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-163b=-104\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{104}{163}\\a=\dfrac{4769}{652}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{4769}{652}\\\dfrac{1}{y}=\dfrac{104}{163}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{652}{4769}\\y=\dfrac{163}{104}\end{matrix}\right.\)(nhận)

c: ĐKXĐ: \(x\ne\pm y\)

Đặt \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{5}{8}\\a-b=-\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=\dfrac{5}{8}-\dfrac{3}{8}=\dfrac{2}{8}=\dfrac{1}{4}\\a+b=\dfrac{5}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=\dfrac{5}{8}-\dfrac{1}{4}=\dfrac{3}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x+y}=\dfrac{1}{4}\\\dfrac{1}{x-y}=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\x-y=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4+\dfrac{8}{3}=\dfrac{20}{3}\\x+y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=4-x=4-\dfrac{10}{3}=\dfrac{2}{3}\end{matrix}\right.\left(nhận\right)\)

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:

\(\left\{{}\begin{matrix}4a+5b=-2\\-5a+3b=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20a+25b=-10\\-20a+12b=84\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}37b=84-10=74\\4a+5b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=-3\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\9x+3y=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\y=\dfrac{1}{2}-3x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{1}{2}-3\cdot\dfrac{7}{66}=\dfrac{1}{2}-\dfrac{7}{22}=\dfrac{4}{22}=\dfrac{2}{11}\end{matrix}\right.\)

e: ĐKXĐ:\(x\ne y-2;x\ne-y+1\)

Đặt x-y+2=a; x+y-1=b

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}\dfrac{7}{a}-\dfrac{5}{b}=4,5\\\dfrac{3}{a}+\dfrac{2}{b}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{14}{a}-\dfrac{10}{b}=9\\\dfrac{15}{a}+\dfrac{10}{b}=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{29}{a}=29\\\dfrac{3}{a}+\dfrac{2}{b}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\\dfrac{2}{b}=4-\dfrac{3}{a}=4-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

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