a: \(x^3-x^2-14x+24\)

\(=x^3+x-12-x^2-15x+36\)

=>\(\left(x^3-x^2-14x+24\right):\left(x^3+x-12\right)=1+\frac{-x^2-15x+36}{x^3+x-12}\)

Để dư là 0 thì \(-x^2-15x+36=0\)

=>\(x^2+15x-36=0\) (1)

\(\Delta=15^2-4\cdot1\cdot\left(-36\right)=225+144=369>0\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-15-\sqrt{369}}{2\cdot1}=\frac{-15-3\sqrt{41}}{2}\\ x=\frac{-15+3\sqrt{41}}{2}\end{array}\right.\)

b: \(x^5+4x^3+3x^2-5x+15\)

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

=>\(\frac{x^5+4x^3+3x^2-5x+15}{x^3-x+3}=x^2+5\)

=>Đây là phép chia hết

c: \(2x^4+2x^3+3x^2-5x-20\)

\(=2x^4+2x^3+8x^2-5x^2-5x-20=\left(x^2+x+4\right)\left(2x^2-5\right)\)

=>\(\frac{2x^4+2x^3+3x^2-5x-20}{x^2+x+4}=2x^2-5\)

d: \(2x^4-14x^3+19x^2-20x+9\)

\(=2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\)

\(=\left(x^2-4x+1\right)\left(2x^2-6x-7\right)-42x+16\)

=>\(\frac{2x^4-14x^3+19x^2-20x+9}{x^2-4x+1}=2x^2-6x-7\) dư -42x+16

để dư bằng 0 thì -42x+16=0

=>-42x=-16

=>\(x=\frac{16}{42}=\frac{8}{21}\)

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

b, \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)=-\frac{3}{2}x^3+x^2-\frac{1}{2}x\)

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

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

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

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

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

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

c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)

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

`@` `\text {Ans}`

`\downarrow`

Bài 1:

a) \(3x\left(5x^2-2x+1\right)\)

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

b) \(\left(x^2-1\right)\left(x^2+2x\right)\)

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

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

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

Bài 2:

a) \(3x^2=2x\)

\(\Leftrightarrow3x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)

b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)

4x^3-3x^2 +1 x^2+2x-1 4x 4x^3+8x^2-4x - -11x^2+4x+1 -11 -11x^2-22x+11 - 26x-10

OLM chỉ có phần chụp ảnh cho CTV

Lưu ý bạn cố phải viết thẳng hàng vì OLM ko viết đc

a: \(=2x^2-4x-6x+12-2x^2+10x=12\)