2: Thay x=0 và y=2 vào y=mx+2, ta được:
\(m\cdot0+2=2\)
=>2=2(đúng)
Vậy: (d) luôn đi qua A(0;2)
1:
ĐKXĐ: x<>3 và y>=-3
\(\left\{{}\begin{matrix}\dfrac{1}{x-3}+3\sqrt{y+3}=7\\\dfrac{5}{x-3}-2\sqrt{y+3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{x-3}+15\sqrt{y+3}=35\\\dfrac{5}{x-3}-2\sqrt{y+3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}17\sqrt{y+3}=34\\\dfrac{5}{x-3}-2\sqrt{y+3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y+3}=\dfrac{34}{17}=2\\\dfrac{5}{x-3}=1+2\sqrt{y+3}=1+2\cdot2=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+3=4\\x-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4\end{matrix}\right.\left(nhận\right)\)
