\(x(x-1)(x+1)-(x-3)(x^2+2x+9)\)

\(=x\left(x^2-1\right)-\left[x\left(x^2+2x+9\right)-3\left(x^2+2x+9\right)\right]\)

\(=x^3-x-\left(x^3+2x^2+9x-3x^2-6x-27\right)\)

\(=x^3-x-\left(x^3-x^2+3x-27\right)\)

\(=x^3-x-x^3+x^2-3x+27\)

\(=x^2-4x+27\)

#\(Toru\)

Đề đâu ah.

$D\,=2x(10x^2-5x-2)-5x(4x^2-2x-1)\\\quad =20x^3-10x^2-4x-20x^3+10x^2+5x\\\quad =(20x^3-20x^3)+(-10x^2+10x^2)+(-4x+5x)\\\quad =x$

Thay $x=-5$ vào $D=x$

$\Rightarrow D=-5$

Vậy $D=-5$ với $x=-5$

Ta có: \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)

\(=20x^3-10x^2-4x-20x^2+10x^2+5x\)

=x=-5

bài này dễ mà

Đề bài yêu cầu gì?

rút gọn à banj

đúng rồi á

9x² - 4 = (2x - 1)(3x + 2)

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

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

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

=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

\(2x-3\text{​​}\text{ ⋮ }x-1\)

\(\Rightarrow2\left(x-1\right)-1\text{ ⋮ }x-1\)

\(\text{mà 2(x-1) ⋮ x-1 }\)\(\Rightarrow1\text{ ⋮ }x-1\)

\(\Rightarrow\)\(x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Vậy \(x\in\left\{0;2\right\}\)