+) Tính giá trị của  x² + 4x - 1 tại x = -2 + \(\sqrt{5}\)

=> (-2 + \(\sqrt{5}\)) ² + 4.(-2 + \(\sqrt{5}\)) - 1 = 4 - 4\(\sqrt{5}\) + 5 - 8 + 4\(\sqrt{5}\) - 1   = 0 

Vậy x² + 4x - 1  = 0 tại x = -2 + \(\sqrt{5}\)

+) A = 3x³.(x² + 4x  - 1 ) - 5x³ - 23x² - 7x + 1

       = 3x³.(x² + 4x  - 1 ) - 5x.(x² + 4x - 1) - 3x² - 12x + 1

      = (3x³ - 5x).(x² + 4x  - 1 ) - 3.(x² + 4x -1) - 2 =  (3x³ - 5x - 3).(x² + 4x  - 1 )  - 2

Vậy tại x = - 2 + \(\sqrt{5}\) thì A = - 2 

+) A =  (3x³ - 5x - 3).(x² + 4x  - 1 )  - 2 chia cho (x² + 4x  - 1 ) dư - 2

\(x+y=14\) ; \(xy=\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=1\)

\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2.1=194\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=194^2-2.1^2=37634\)

\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-\left(xy\right)^3\left(x+y\right)=2702.37634-1^3.14=...\)

a) ĐK:\(x\ge0;x\ne9\)

\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b)\(P=-\dfrac{3}{\sqrt{x}+3}\) 

Có \(\sqrt{x}+3\ge3;\forall x\ge0\)

\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)

a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)

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

b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)

\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)

Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)