f: Ta có: \(x\left(2x-9\right)-4x+18=0\)

\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)

g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)

\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)

f. x(2x - 9) - 4x + 18 = 0

<=> x(2x - 9) - 2(2x - 9) = 0

<=> (x - 2)(2x - 9) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)

g. 4x(x - 1000) - x + 1000 = 0

<=> 4x(x - 1000) - (x - 1000) = 0

<=> (4x - 1)(x - 1000) = 0

<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)

h. 2x(x - 4) - 6x²(-x + 4) = 0

<=> 2x(x - 4) + 6x²(x - 4) = 0

<=> (2x + 6x²)(x - 4) = 0

<=> 2x(1 + 3x)(x - 4) = 0

<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)

i. 2x(x - 3) + x² - 9 = 0

<=> 2x(x - 3) + (x - 3)(x + 3) = 0

<=> (2x + x + 3)(x - 3) = 0

<=> (3x + 3)(x + 3) = 0

<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

j. 9x - 6x² + x³ = 0

<=> x(9 - 6x + x²) = 0

<=> x(3 - x)² = 0

<=> \(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

 4x(x – 1000) – x + 1000 = 0

(4x-1)(x-1000)                 =0

⇔x=1/4 hoặc  1000

\(=12x^3-10x+18x^2-15\)

\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\\ =\left[\left(6x^3-6x^2\right)-\left(x^2-x\right)-\left(2x-2\right)\right]:\left(2x+1\right)\\ =\left[\left(x-1\right)\left(6x^2-x-2\right)\right]:\left(2x+1\right)\\ =\left\{\left(x-1\right)\left[\left(6x^2+3x\right)-\left(4x+2\right)\right]\right\}:\left(2x+1\right)\\ =\left[\left(x-1\right)\left(3x-2\right)\left(2x+1\right)\right]:\left(2x+1\right)\\ =\left(x-1\right)\left(3x-2\right)\)

\(\left(6x^2-5\right)\left(2x+3\right)=2x\left(6x^2-5\right)+3\left(6x^2-5\right)=12x^3-10x+18x^2-15\)

Câu 1: A

Caau 2: B

\(=\left(6x^2-9x+10x-15\right):\left(2x-3\right)\\ =\left[3x\left(2x-3\right)+5\left(2x-3\right)\right]:\left(2x-3\right)\\ =3x+5\)

\(\dfrac{6x^2-2x+1}{x-1}\)

\(=\dfrac{6x^2-6x+4x-4+5}{x-1}=6x+4+\dfrac{5}{x-1}\)

\(6x^2-9x-4\left(2x-3\right)=0\)

\(\Leftrightarrow3x\left(2x-3\right)-4\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{3}{2};\dfrac{4}{3}\right\}\)

\(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)

\(\Rightarrow6x^2-6x^2+5x+6-1=0\)

\(\Rightarrow5x=-5\Rightarrow x=-1\)

\(\Rightarrow6x^2-\left(6x^2-5x-6\right)-1=0\\ \Rightarrow5x+5=0\\ \Rightarrow x=-1\)

\(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)

\(\Leftrightarrow6x^2-6x^2-4x+9x+6-1=0\)

\(\Leftrightarrow x=-1\)

( 36x +12x5 - 8x4 + 10x3 - 6x2 + 2x - 1) : (x4 + 4x3 - 3x2 + 2x - 1)

=12x-56

Dư 120x³-98x²+112x+55

(Mình tính ra nháp r hí hí)