Giải hệ phương trình \(\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{6}=-1\\2x+5y=-9\end{matrix}\right.\) bằng phương pháp thế. \(\left(x,y\right)=\left(-\dfrac{21}{13};-\dfrac{15}{13}\right)\) \(\left(x,y\right)=\left(-\dfrac{25}{13};-\dfrac{10}{13}\right)\) \(\left(x,y\right)=\left(-\dfrac{30}{17};-\dfrac{6}{17}\right)\) \(\left(x,y\right)=\left(-\dfrac{15}{17};-\dfrac{12}{17}\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{6}=-1\\2x+5y=-9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x+y=-6\\2x+5y=-9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-6-3x\\2x+5\left(-6-3x\right)=-9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-6-3x\\-13x=21\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-6-3x\\x=-\dfrac{21}{13}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{21}{13}\\y=-\dfrac{15}{13}\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}x+\sqrt{3}y=0\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\) bằng phương pháp thế. \(\left(x,y\right)=\left(3+\sqrt{3},-1-\sqrt{3}\right)\) \(\left(x,y\right)=\left(3-\sqrt{3},1+\sqrt{3}\right)\) \(\left(x,y\right)=\left(-3+\sqrt{3},1-\sqrt{3}\right)\) \(\left(x,y\right)=\left(3+2\sqrt{3},1+\sqrt{3}\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}x+\sqrt{3}y=0\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{3}y\\-y=1+\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{3}\left(-1-\sqrt{3}\right)\\y=-1-\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+\sqrt{3}\\y=-1-\sqrt{3}\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\) bằng phương pháp thế. \(\left(x,y\right)=\left(\dfrac{4+\sqrt{3}}{3};-\dfrac{1}{3}\right)\) \(\left(x,y\right)=\left(\dfrac{4-\sqrt{3}}{3};-\dfrac{\sqrt{3}}{3}\right)\) \(\left(x,y\right)=\left(\dfrac{1-\sqrt{3}}{3};-\dfrac{1+\sqrt{3}}{3}\right)\) \(\left(x,y\right)=\left(\dfrac{4-2\sqrt{3}}{3};\dfrac{\sqrt{3}}{3}\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\x+\left(\sqrt{3}+1\right)\left[\left(\sqrt{3}-1\right)x-\sqrt{3}\right]=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\x+2x-\sqrt{3}\left(\sqrt{3}+1\right)=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{3}-1\right)x-\sqrt{3}\\3x=4+\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{3}-\sqrt{3}\\x=\dfrac{4+\sqrt{3}}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Xác định a, b để hệ phương trình \(\left\{{}\begin{matrix}-2ax+5by=3\\3ax+4by=30\end{matrix}\right.\) có nghiệm \(\left(x,y\right)=\left(-2;3\right)\). \(a=-3,b=1\) \(a=3;b=-1\) \(a=-1,b=-1\) \(a=2;b=-2\) Hướng dẫn giải: Thay \(\left(x,y\right)=\left(-2;3\right)\) vào hệ phương trình \(\left\{{}\begin{matrix}-2ax+5by=3\\3ax+4by=30\end{matrix}\right.\) ta có: \(\left\{{}\begin{matrix}4a+15b=3\\-6a+12b=30\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4a+15b=3\\-a+2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=1\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=-4\\4\left(x+y\right)+5\left(x-y\right)=-2\end{matrix}\right.\) bằng phương pháp thế. \(\left(x,y\right)=\left(\dfrac{1}{2},\dfrac{13}{2}\right)\) \(\left(x,y\right)=\left(\dfrac{3}{2},2\right)\) \(\left(x,y\right)=\left(1,\dfrac{13}{2}\right)\) \(\left(x,y\right)=\left(-\dfrac{3}{2},-2\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=-4\\4\left(x+y\right)+5\left(x-y\right)=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-y=-4\\9x-y=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5x+4\\9x-5x-4=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5x+4\\4x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{13}{2}\end{matrix}\right.\)
Từ hệ phương trình \(\left\{{}\begin{matrix}3x+4y=-7\\5x+2y=-1\end{matrix}\right.\) ta có thể suy ra hệ phương trình nào trong số các hệ phương trình dưới đây: (Được chọn hai đáp án đúng) \(\left\{{}\begin{matrix}15x+20y=-35\\15x+6y=-3\end{matrix}\right.\) \(\left\{{}\begin{matrix}15x+20y=-7\\15x+6y=-1\end{matrix}\right.\) \(\left\{{}\begin{matrix}3x+4y=-7\\10x+4y=-1\end{matrix}\right.\) \(\left\{{}\begin{matrix}3x+4y=-7\\10x+4y=-2\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}x-3y=-6\\x+5y=-7\end{matrix}\right.\) bằng phương pháp cộng đại số. \(\left(x,y\right)=\left(-\dfrac{51}{8},-\dfrac{1}{8}\right)\) \(\left(x,y\right)=\left(-\dfrac{35}{8},-\dfrac{5}{8}\right)\) \(\left(x,y\right)=\left(-\dfrac{1}{8},\dfrac{5}{8}\right)\) \(\left(x,y\right)=\left(-\dfrac{45}{8},\dfrac{1}{8}\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}x-3y=-6\\x+5y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y=-1\\x-3y=-6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{8}\\x=-\dfrac{51}{8}\end{matrix}\right.\) Vậy phương trình có nghiệm \(\left(-\dfrac{51}{8};-\dfrac{1}{8}\right)\)
Giải hệ phương trình \(\left\{{}\begin{matrix}3x+4y=-1\\4x-3y=7\end{matrix}\right.\) bằng phương pháp cộng đại số. \(\left(x,y\right)=\left(1,-1\right)\) \(\left(x,y\right)=\left(1,1\right)\) \(\left(x,y\right)=\left(-1,-1\right)\) \(\left(x,y\right)=\left(1;-2\right)\)
Giải hệ phương trình \(\left\{{}\begin{matrix}-0,5x+1,2y=2,7\\x-4,5y=-7,5\end{matrix}\right.\) bằng phương pháp cộng đại số. \(\left(x,y\right)=\left(-3,1\right)\) \(\left(x,y\right)=\left(3,1\right)\) \(\left(x,y\right)=\left(1,3\right)\) \(\left(x,y\right)=\left(6,-2\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}-0,5x+1,2y=2,7\\x-4,5y=-7,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-x+2,4y=5,4\\x-4,5y=-7,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2,1y=-2,1\\x=4,5y-7,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-3\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{4}=2\\\dfrac{2x}{5}+y=18\end{matrix}\right.\) bằng phương pháp cộng đại số. \(\left(x,y\right)=\left(15,12\right)\) \(\left(x,y\right)=\left(4,5\right)\) \(\left(x,y\right)=\left(-15,20\right)\) \(\left(x,y\right)=\left(-15,-12\right)\) Hướng dẫn giải: \(\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{4}=2\\\dfrac{2x}{5}+y=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x}{3}-y=8\\\dfrac{2x}{5}+y=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{26x}{15}=26\\y=18-\dfrac{2x}{5}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=15\\y=12\end{matrix}\right.\)
