\(\left\{{}\begin{matrix}x=5-y\\y\left(5-y\right)=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\end{matrix}\right.\)

All the paper was blown away by the wind.

Dozens of trees were destroyed by the storm.

Nancy a pay increase was offered (by them)

a) Thay m = -2 vào (P) ta có:

\(y=\left(-2-3\right)x+\left(-2\right).\\ \Leftrightarrow y=-5x-2.\)

i) + \(y=-5x-2.\)

\(Cho\) \(x=0.\Rightarrow y=-2.\)

\(Cho\) \(y=0.\Rightarrow x=\dfrac{-2}{5}.\)

+ \(y=2x^2.\)

1: Thay x=16 vào A, ta được:

\(A=\dfrac{4-1}{4+3}=\dfrac{3}{7}\)

2: \(P=A:B\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}:\dfrac{x-3\sqrt{x}-x-6\sqrt{x}-9+x+11\sqrt{x}+6}{x-9}\)

\(=\dfrac{\sqrt{x}-1}{1}\cdot\dfrac{\sqrt{x}-3}{x+2\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

Bài 2:

\(c,\forall m=0\Leftrightarrow-3x-1=0\Leftrightarrow x=-\dfrac{1}{3}\\ \forall m\ne0\\ \Delta=3^2-4\left(-1\right)\cdot2m^2=8m^2+9>0,\forall m\)

Vậy PT có 2 nghiệm phân biệt với mọi m

Bài 5: 

a: Để đây là hàm số bậc nhất thì m+5<>0

hay m<>-5