e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

\(a,\left(x-1\right)\left(2x-1\right)\)

\(=2x^2-x-2x+1\)

\(=2x^2-3x+1\)

\(b,\left(9x^4+12x^3-15x^2-3x\right):3x\)

\(=3x^3+4x^2-5x-1\)

a) (x + 3y) (2x²y - 6xy²)

= (x + 3y) + 2xy (x - 3y)

= 2xy [(x + 3y) (x - 3y)]

= 2xy (x² - 3y²)

b) (6x⁵y² - 9x⁴y³ + 15x³y⁴) : 3x³y²

= (6x⁵y² : 3x³y²) + (-9x⁴y³ : 3x³y²) + (15x³y⁴ : 3x³y²)

= [(6 : 3) (x⁵ : x³) (y² : y²)] + [(-9 : 3) (x⁴ : x³) (y³ : y²)] + [(15 : 3) (x³ : x³) (y⁴ : y²)]

= 2x² + (-3xy) + 5y²

= 2x² - 3xy + 5y²

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)

b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

\(=y^2-2x^2y^3\)

c: \(=6x-y+2x^2+3y-2x^2+x\)

\(=7x+2y\)

d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

2. Phân tích đa thức thành nhân tử :

b. \(4x^2-9x^2\) =

= \(\left(2x\right)^2-\left(3x\right)^2\)

= \(\left(2x-3x\right)\left(2x+3x\right)\)

c. \(x^2-3xy+5x-15y=\)

= \(\left(x^2-3xy\right)+\left(5x-15y\right)\)

= \(x\left(x-3y\right)+5\left(x-3y\right)\)

= \(\left(x-3y\right)\left(x+5\right)\)

a: \(\frac{11x-3}{3x^2-15x-42}\)

\(=\frac{11x-3}{3\left(x^2-5x-14\right)}=\frac{11x-3}{3\left(x-7\right)\left(x+2\right)}\)

\(=\frac{3\left(11x-3\right)\left(x+1\right)}{3\cdot3\cdot\left(x-7\right)\left(x+2\right)\left(x+1\right)}=\frac{3\left(11x-3\right)\left(x+1\right)}{9\left(x-7\right)\left(x+2\right)\left(x+1\right)}\)

\(\frac{8}{x^2-6x-7}=\frac{8}{x^2-7x+x-7}\)

\(=\frac{8}{\left(x-7\right)\left(x+1\right)}\)

\(=\frac{8\cdot9\cdot\left(x+2\right)}{9\left(x+2\right)\left(x-7\right)\left(x+1\right)}=\frac{72x+144}{9\left(x+2\right)\left(x-7\right)\left(x+1\right)}\)

\(\frac{13x}{9x-63}=\frac{13x}{9\left(x-7\right)}\)

\(=\frac{13x\left(x+2\right)\left(x+1\right)}{9\left(x-7\right)\left(x+2\right)\left(x+1\right)}\)

b: \(\frac{2}{x^2+2x}=\frac{2}{x\left(x+2\right)}\)

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

\(\frac{3x^2-6x}{x^2-2x+4}=\frac{3x\left(x-2\right)}{x^2-2x+4}=\frac{3x\left(x-2\right)\cdot x\left(x+2\right)}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

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

\(\frac{10x^2+28x-8}{x^4+8x}=\frac{10x^2+28x-8}{x\left(x^3+8\right)}=\frac{10x^2+28x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)