\(M=\frac{x^2+2x-9}{x-3}\)

\(=\frac{x^2-6x+9+8x-24+6}{x-3}\)

\(=\frac{\left(x-3\right)^2+8\left(x-3\right)+6}{x-3}\)

\(=x-3+8+\frac{6}{x-3}\)

Do \(x>3\Rightarrow x-3>0\)

Áp dụng BĐT Cauchy , ta có : 

\(x-3+\frac{6}{x-3}\ge2\sqrt{\left(x-3\right).\frac{6}{x-3}}=2\sqrt{6}\)

\(\Rightarrow M=x-3+\frac{6}{x-3}+8\ge2\sqrt{6}+8\)

\(\Rightarrow M\ge\sqrt{24}+8\)

Dấu " = " xảy ra \(\Leftrightarrow x-3=\frac{6}{x-3}\Leftrightarrow\left(x-3\right)^2=6\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{6}\\x-3=-\sqrt{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{6}\left(TM\right)\\x=3-\sqrt{6}\left(L\right)\end{cases}}}\)

Vậy Min M là : \(\sqrt{24}+8\Leftrightarrow x=3+\sqrt{6}\)

Ta có : 

2(xy - x^2 - y + 1008) = y^2 + 2018

<=> 2xy - 2x^2 -  2y + 2016 = y^2 + 2018

<=> 2xy - 2x^2 - 2y = y^2 + 2

<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0 

<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0 

<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0

<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0 

<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0 

Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0

<=> x = y/2 ; y = -2

<=> x = -1 ; y = -2

Vậy x = -1 ; y = -2

Ta có : 

2(xy - x^2 - y + 1008) = y^2 + 2018

<=> 2xy - 2x^2 -  2y + 2016 = y^2 + 2018

<=> 2xy - 2x^2 - 2y = y^2 + 2

<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0 

<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0 

<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0

<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0 

<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0 

Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0

<=> x = y/2 ; y = -2

<=> x = -1 ; y = -2

Vậy x = -1 ; y = -2

Ta có : 

x/x^2 + x + 1 = -2/3

<=> -2x^2 - 2x - 2 = 3x

<=> -2x^2 - 5x - 2 = 0 

<=> -2(x^2 + 5/2x + 1) = 0 

<=> x^2 + 5/2x + 1 = 0

<=> x^2 + 2x.5/4 + 25/16 - 9/16 = 0 

<=> (x+5/4)^2 = 9/16

<=> x + 5/4 = 3/4 hoặc x + 5/4 = -3/4

<=> x = -1/2 hoặc x = -2

Sau đấy thay vào ( easy ) 

x/x^2 + x + 1 = -2/3

<=> -2x^2 - 2x - 2 = 3x

<=> -2x^2 - 5x - 2 = 0 

<=> -2(x^2 + 5/2x + 1) = 0 

<=> x^2 + 5/2x + 1 = 0

<=> x^2 + 2x.5/4 + 25/16 - 9/16 = 0 

<=> (x+5/4)^2 = 9/16

<=> x + 5/4 = 3/4 hoặc x + 5/4 = -3/4

<=> x = -1/2 hoặc x = -2

a)\(x\ne1;x\ne-1\)

\(A=\left(\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right):\left(\frac{x^2+1-2x}{x^2+1}\right)\)

\(A=\left(\frac{1}{x-1}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right).\frac{x^2+1}{x^2+1-2x}\)

\(A=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\frac{x^2+1}{x^2+1-2x}\)

\(A=\frac{1}{x-1}\)

Đặt \(f\left(x\right)=x^2+3x+a\)

    \(g\left(x\right)=x-5\)

Vì f(x) chia hết cho g(x) nên

\(f\left(x\right)=q\left(x\right).g\left(x\right)\)

\(\Leftrightarrow x^2+3x+a=q\left(x\right).\left(x-5\right)\)

Tại x = 5 thì

\(5^2+3.5+a=0\)

\(\Leftrightarrow a=-40\)

Vậy a = -40

Ta có:\(x^2+3x+a=\left(x+5\right)\left(x+8\right)\)  dư a-40

Vậy để \(x^2+3x+a⋮x-5\)Thì a-40=0=>z=40

Vậy để \(x^2+3x+a⋮x-5\)thì a=40