`#3107.101107`

`4x = 5y => x/5 = y/4`

Đặt `x/5 = y/4 = k`

`=> x = 5k; y = 4k`

Ta có: `x^2 - y^2 = 1`

`=> (5k)^2 - (4k)^2 = 1`

`=> 25k^2 - 16k^2 = 1`

`=> 9k^2 = 1`

`=> k^2 = 1 \div 9`

`=> k^2 = 1/9`

`=> k^2 = (+-1/3)^2`

`=> k = +-1/3`

Với `k = 1/3`

`=> x = 1/3*5 = 5/3; y = 1/3*4 = 4/3`

Với `k = -1/3`

`=> x = -1/3*5 = -5/3; y = -1/3*4 = -4/3.`

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\)

Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\\end{matrix}\right.\)

Thay vào \(x^2-y^2=1\)

\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)

\(\Leftrightarrow25k^2-16k^2=1\)

\(\Leftrightarrow9k^2=1\)

\(\Leftrightarrow k^2=\dfrac{1}{9}\)

\(\Leftrightarrow k=\pm\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=5k=5.\dfrac{1}{3}=\dfrac{5}{3}\\y=4k=4.\dfrac{1}{3}=\dfrac{4}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=5k=5.\left(-\dfrac{1}{3}\right)=-\dfrac{5}{3}\\y=4k=4.\left(-\dfrac{1}{3}\right)=-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)

\(a^2+b^2+6ab+2=2a+3b\Rightarrow\left(a+b\right)^2-3\left(a+b\right)+2=-a\left(4b+1\right)\le0\)

\(\Rightarrow\left(a+b-1\right)\left(a+b-2\right)\le0\Rightarrow1\le a+b\le2\)

\(a^2+b^2+6ab+2=2a+3b\Rightarrow4ab=-\left(a+b\right)^2+2a+3b-2\)

\(-P=\dfrac{6a+5b+4ab+7}{a+b+1}=\dfrac{6a+5a+7-\left(a+b\right)^2+2a+3b-2}{a+b+1}\)

\(=\dfrac{-\left(a+b\right)^2+8\left(a+b\right)+5}{a+b+1}\)

Tới đây có thể giải theo lớp 9 (tách thành tích hoặc bình phương) hoặc làm theo lớp 12 (khảo sát hàm \(f\left(x\right)=\dfrac{-x^2+8x+5}{x+1}\) trên \(\left[1;2\right]\)). Cả 2 việc đều dễ dàng cả

\(-P=6-\dfrac{\left(x-1\right)^2}{x+1}=\dfrac{17}{3}+\dfrac{\left(3x-1\right)\left(2-x\right)}{3\left(x+1\right)}\)