Câu 1:
a: \(A=3\sqrt{2}+5\sqrt{2}-2\cdot3\sqrt{2}=2\sqrt{2}\)
b: \(P=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+4}=\dfrac{1}{\sqrt{x}-2}\)
c: Vì y=ax+b//y=2x+1 nên a=2
Vậy: y=2x+b
Thay x=-1 và y=2 vào y=2x+b, ta được:
b-1=2
hay b=3
Câu 2
phần 1
\(\left\{{}\begin{matrix}2x+3y=-2\\5x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2-3y}{2}\\\dfrac{-10-15y}{2}+2y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2-3y}{2}\\\dfrac{-10-11y}{2}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2-3y}{2}\\-22-11y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2-3\left(-2\right)}{2}=2\\y=-2\end{matrix}\right.\)
.
