\(x^3+9x=0\)

<=> \(x\left(x^2+9\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)

<=> \(x=0\)

\(9x^2-4-2\left(3x-2\right)^2=0\)

<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)

<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)

<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)

<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)

<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)

<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)

<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)

<=> \(3\left(3x-2\right)\left(2-x\right)=0\)

<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)

\(\left(x^3-x^2\right)-4x+8x-4=0\)

<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)

<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)

<=> \(\left(x-1\right)\left(x^2+4\right)=0\)

<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)

<=> \(x=1\)

\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)

<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)

<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)

<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)

<=> \(5x-2+3x-6=-4\)

<=> \(8x-8=-4\)

<=> \(8\left(x-1\right)=-4\)

<=> \(x-1=-\frac{1}{2}\)

<=> \(x=-\frac{3}{2}\)

a) (x + 2)(x² + 3x + 1)

= x.x² + x.3x + x.1 + 2.x² + 2.3x + 2.1

= x³ + 3x² + x + 2x² + 6x + 2

= x³ + 5x² + 7x + 2

b) (2x³ + 10x² + 9x + 4) : (x + 4)

= (2x³ + 8x² + 2x² + 8x + x + 4) : (x + 4)

= [(2x³ + 8x²) + (2x² + 8x) + (x + 4)] : (x + 4)

= [2x²(x + 4) + 2x(x + 4) + (x + 4)] : (x + 4)

= (x + 4)(2x² + 2x + 1) : (x + 4)

= 2x² + 2x + 1

a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)

\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)

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

c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

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

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

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)

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

\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)

\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)

\(=x^2-2x-5\)

\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)

\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)

\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)

\(=2x-3\)

\(\frac{\left(8x^3-y^3\right)\left(4x^2-y^2\right)}{\left(2x+y\right)\left(4x^2-4xy+y^2\right)}\)

\(=\frac{\left(2x-y\right)\left(4x^2+2xy+y^2\right)\left(2x-y\right)\left(2x+y\right)}{\left(2x+y\right)\left(2x-y\right)^2}\)

\(=4x^2+2xy+y^2\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{5\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3x}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)

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

b) Ta có: \(9x^2-10x+1=0\)

\(\Leftrightarrow\left(9x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào P, ta được:

\(P=\dfrac{1}{3-1}=\dfrac{1}{2}\)

c) Thay \(x=8-2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{8-2\sqrt{7}}{3\left(\sqrt{7}-1\right)-1}=\dfrac{8-2\sqrt{7}}{3\sqrt{7}-4}\)

\(=\dfrac{-10+16\sqrt{7}}{47}\)

a)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{3x-2\sqrt{x}-1-3\sqrt{x}+4+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

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

b) Từ phương trình suy ra \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

Vói x=1 

\(P=\dfrac{1}{3\sqrt{1}-1}=\dfrac{1}{2}\)

Với x= 1/9

\(P=\dfrac{\dfrac{1}{9}}{3\sqrt{\dfrac{1}{9}}-1}\) không có nghiệm