làm gấp giúp mik vs ak
15: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\y>=0\end{matrix}\right.\)
Đặt \(\sqrt{x}=a\left(a>=0\right);\sqrt{y}=b\left(b>=0\right)\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}3a+2b=16\\2a-3b=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+2b=16\\2a=3b-11\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{3b-11}{2}\\3\cdot\dfrac{3b-11}{2}+2b=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3b-11}{2}\\\dfrac{9b-33}{2}+\dfrac{4b}{2}=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{3b-11}{2}\\13b-33=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13b=65\\a=\dfrac{3b-11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=5\\a=\dfrac{3\cdot5-11}{2}=\dfrac{15-11}{2}=2\end{matrix}\right.\left(nhận\right)\)
=>\(\left\{{}\begin{matrix}x=2^2=4\\y=5^2=25\end{matrix}\right.\left(nhận\right)\)
10: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{3}{x}-\dfrac{2}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\3\cdot\dfrac{1}{x}-\dfrac{2}{y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\3\left(3-\dfrac{1}{y}\right)-\dfrac{2}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\9-\dfrac{5}{y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{y}=9+1=10\\\dfrac{1}{x}=3-\dfrac{1}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\\dfrac{1}{x}=3-1:\dfrac{1}{2}=3-2=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\left(nhận\right)\)
11: ĐKXĐ: y<>0
\(\left\{{}\begin{matrix}x+\dfrac{1}{y}=-\dfrac{1}{2}\\2x-\dfrac{3}{y}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}-\dfrac{1}{y}\\2\left(-\dfrac{1}{2}-\dfrac{1}{y}\right)-\dfrac{3}{y}=-\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{y}-\dfrac{1}{2}\\-1-\dfrac{2}{y}-\dfrac{3}{y}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{y}-\dfrac{1}{2}\\1+\dfrac{5}{y}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{y}-\dfrac{1}{2}\\\dfrac{5}{y}=\dfrac{7}{2}-1=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-\dfrac{1}{2}-\dfrac{1}{2}=-1\end{matrix}\right.\left(nhận\right)\)
13: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{1}{y-2}=\dfrac{5}{x+1}-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{y-2}=\dfrac{5}{x+1}-3\\\dfrac{1}{x+1}+2\left(\dfrac{5}{x+1}-3\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-2}=\dfrac{5}{x+1}-3\\\dfrac{11}{x+1}=11\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1=1\\\dfrac{1}{y-2}=\dfrac{5}{1}-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\left(nhận\right)\)
14: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-\dfrac{y}{2}\\x\ne2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{2x+y}+\dfrac{1}{x-2y}=\dfrac{5}{8}\\\dfrac{1}{2x+y}-\dfrac{1}{x-2y}=-\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x+y}+\dfrac{1}{x-2y}=\dfrac{5}{8}\\\dfrac{1}{x-2y}=\dfrac{1}{2x+y}+\dfrac{3}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x-2y}=\dfrac{1}{2x+y}+\dfrac{3}{8}\\\dfrac{1}{2x+y}+\dfrac{1}{2x+y}+\dfrac{3}{8}=\dfrac{5}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x+y}=\dfrac{2}{8}\\\dfrac{1}{x-2y}=\dfrac{1}{2x+y}+\dfrac{3}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+y=8\\\dfrac{1}{x-2y}=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{4}{8}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=8\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=16\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=18\\2x+y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{18}{5}\\y=8-2x=8-2\cdot\dfrac{18}{5}=8-\dfrac{36}{5}=\dfrac{4}{5}\end{matrix}\right.\left(nhận\right)\)
