ĐK: x,y ≥ 0

\(\sqrt{x}+\sqrt{y}-1=\sqrt{x+y+3}\)

\(\Rightarrow x+y+1+2\sqrt{xy}-2\sqrt{x}-2\sqrt{y}=x+y+3\)

\(\Rightarrow\sqrt{xy}-1=\sqrt{x}+\sqrt{y}\)

\(\Rightarrow xy+1-2\sqrt{xy}=x+y-2\sqrt{xy}\)

\(\Rightarrow x-xy+y-1=0\)

\(\Rightarrow-x\left(y-1\right)+\left(y-1\right)=0\)

\(\Rightarrow\left(1-x\right)\left(y-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Ta có: |x² + x + 13| = |(x + 1/2)² + 51/4| ≥ 51/4 ∀x

|x² + x - 9| = |(x+1/2)² - 37/4| ≥ 37/4 ∀x

⇒ A ≥ 51/4 + 37/4 = 22 ∀x

Dấu "=" xảy ra khi x = -1/2

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{a}=2\\\dfrac{c+a}{b}=2\\\dfrac{a+b}{c}=2\end{matrix}\right.\)

\(\Rightarrow A=2024-2+2-2=2022\)

1. did you meet - met

2. didn't eat - ate

Theo tính chất của dãy tỉ số bằng nhau:

\(x=\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{b+c+a+c+a+b}{a+b+c}=2\)

⇒ A = (x² - x + 1)¹⁰ = (2² - 2 + 1)¹⁰ = 3¹⁰