Ta có: 

\(2x^3-5x^2+6x-15\)

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

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

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

\(\Rightarrow\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)=x^2+3\)

= x + 3 nha

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

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

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

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

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

\(\Leftrightarrow x^2+x-8=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)

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

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

2) \(\left(x+3\right)\left(5x-15\right)+2x-6=0\)

\(\Leftrightarrow\left(x+3\right).5.\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[5\left(x+3\right)+2\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+17\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)

câu 1 phải là \(\left(x-5\right)\left(x+9\right)+6x=30\)

\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6x-30=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+9+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-15\end{matrix}\right.\)

tìm hả bn

\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)

c: =(x+5)(x-3)

d: =(x-1)(7x+1)

\(a,5x^2-2x-3=\left(5x^2-5x\right)+\left(3x-3\right)=5x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(5x+3\right)\\ b,2x^2-3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\\ c,x^2+2x-15=\left(x^2-3x\right)+\left(5x-15\right)=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\\ d,7x^2-6x-1=\left(7x^2-7x\right)+\left(x-1\right)=7x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(7x+1\right)\)

b. 6x(x - 5) - x(6x + 3)

= x(6x - 30) - x(6x + 3)

= x(6x - 30 - 6x - 3)

= x(-33)

= -33x

\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)