\(g,\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\x-15y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=15y+16\\30y+32+10y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=15y+8\\40y=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{33}{40}\cdot15+8=\dfrac{29}{8}\\y=-\dfrac{33}{40}\end{matrix}\right.\)
\(h,\Leftrightarrow\left(x-15\right)\left(y+2\right)=\left(x+15\right)\left(y-1\right)\\ \Leftrightarrow xy+2x-15y-30=xy-x+15y-15\\ \Leftrightarrow3x=30y+15\Leftrightarrow x=10y+5\\ \text{Ta có }PT\left(1\right)\Leftrightarrow xy+2x-15y-30=xy\Leftrightarrow2x-15y=30\\ \Leftrightarrow2\left(10y+5\right)-15y=30\\ \Leftrightarrow y=4\Leftrightarrow x=45\)
\(i,\Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\\sqrt{5}\left(5\sqrt{5}-2y\right)+y=5+2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\25-2\sqrt{5}y+y=5+2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\y\left(1-2\sqrt{5}\right)=20-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}-2y\\y=\dfrac{20-2\sqrt{5}}{1-2\sqrt{5}}=\dfrac{2\sqrt{5}\left(2\sqrt{5}-1\right)}{-\left(2\sqrt{5}-1\right)}=-2\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{5}+4\sqrt{5}=9\sqrt{5}\\y=-2\sqrt{5}\end{matrix}\right.\)
Câu i sửa: \(PT\left(2\right)\Leftrightarrow y=5+2\sqrt{5}-\sqrt{5}x\)
Thế vào \(PT\left(1\right)\Leftrightarrow x+10+4\sqrt{5}-2\sqrt{5}x=5\sqrt{5}\)
\(\Leftrightarrow x\left(1-2\sqrt{5}\right)=\sqrt{5}-10\\ \Leftrightarrow x=\dfrac{\sqrt{5}-10}{1-2\sqrt{5}}=\dfrac{\sqrt{5}\left(1-2\sqrt{5}\right)}{1-2\sqrt{5}}=\sqrt{5}\\ \Leftrightarrow y=5+2\sqrt{5}-5=2\sqrt{5}\)
Lập bảng và giải bằng hpt giúp e ạ
