A = 5(x + 3)(x - 3) + (2x + 3)³ + (x - 6)²

A = 5(x + 3)(x - 3) + 4x² + 12x + 9 + x² - 12x + 36

A = 5x² - 45x + 4x² + 12x + 9 + x² - 12x + 36

A = 10x² (1)

Thay x = -1/5 vào (1), ta có:

A = 10x² = 10.(-1/5)² = 2/5

A = 2/5

Vậy:...

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

1) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-9\ne0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}\)2) Để A=\(\dfrac{5}{6}\) thì \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}=\dfrac{5}{6}\Leftrightarrow\left(\sqrt{x}+1\right)6=\left(\sqrt{x}+3\right)5\Leftrightarrow6\sqrt{x}+6=5\sqrt{x}+15\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

1. Ta có:

\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)}{3\left(x-9\right)}+\dfrac{1}{3}\)

\(=\dfrac{2x-6\sqrt{x}}{3\left(x-9\right)}+\dfrac{x-9}{3\left(x-9\right)}\)

\(=\dfrac{3x-6\sqrt{x}-9}{3x-27}\)

\(=\dfrac{x-2\sqrt{x}-3}{x-9}\)

a: \(=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

b: Khi x=14-6căn 5 thì \(A=\dfrac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\dfrac{58-2\sqrt{5}}{11}\)

a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)

Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : 

\(B=\frac{2+5}{2-3}=-7\)

b, Ta có : Với \(x\ge0;x\ne9\)

\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)

Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

Ta có : \(Q=\frac{x^6-3x^5+3x^4-x^3+2020}{x^6-x^3-3x^2-3x+2020}\)

=> \(Q=\frac{\left(x^6-x^5-x^4\right)+\left(-2x^5+2x^4+2x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(-x^3+x^2+x\right)+\left(x^2-x-1\right)+2021}{\left(x^6-x^5-x^4\right)+\left(x^5-x^4-x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(2x^3-2x^2-2x\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4\left(x^2-x-1\right)-2x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}{x^4\left(x^2-x-1\right)+x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4.0-2x^3.0+2x^2.0-x.0+0+2021}{x^4.0+x^3.0+2x^2.0+0+2021}\)

=> \(Q=\frac{2021}{2021}=1\)

+) \(P=3x-6-3x+5=-1\)

+) \(Q=2x-8+3x+8=5x\)

+) R bạn xem lại điều kiện

Ta có: P=|3x-6|-3x+5

=3x-6-3x+5

=-1

Ta có: Q=|8-2x|+3x+8

=2x-8+3x+8

=5x