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b: \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=-6\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=5-2x=5-12=-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+3y=-4\\5x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4-3y\\5\left(-4-3y\right)-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4-3y\\-20-15y-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4-3y\\-20-23y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4-3\left(-1\right)\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=3\\x+2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3+y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3-2\\y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=3\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy hpt có nghiệm (x;y) = (1;2)
\(\left(2\right)\Leftrightarrow\left|x-1\right|=3-3y\)
Thay vào \(\left(1\right)\Leftrightarrow3-3y+\left|y-2\right|=1\Leftrightarrow\left|y-2\right|=3y-2\)
\(\Leftrightarrow\left[{}\begin{matrix}y-2=3y-2\left(y\ge2\right)\\2-y=3y-2\left(y< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\left(tkm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
Với \(y=1\Leftrightarrow\left|x-1\right|=3-3=0\Leftrightarrow x=1\)
Vậy \(\left(x;y\right)=\left(1;1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{135}{x}-\dfrac{63}{y}=81\\\dfrac{28}{x}+\dfrac{63}{y}=245\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{163}{x}=326\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{9}{y}=35-\dfrac{4}{x}=35-8=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x+2y=-2\\3x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\3x+2y-3x+2y=-2+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\4y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2.\dfrac{1}{4}=-2\\y=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{6}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-5x+2y=4\\6x-3y=-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\15x-6y-12x+6y=-12-\left(-14\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\3x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15.\dfrac{2}{3}-6y=-12\\x=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\-5x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\2y-\dfrac{10}{3}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)