Mọi người giúp em với nhé

Giải thích các bước giải:

Bài 1:

a) $2x(3x^2 - 5x + 3)$

$\begin{aligned}
&= 2x \cdot 3x^2 - 2x \cdot 5x + 2x \cdot 3 \\
&= 6x^3 - 10x^2 + 6x
\end{aligned}$

b) $-2x(x^2 + 5x - 3)$

$\begin{aligned}
&= -2x \cdot x^2 + (-2x) \cdot 5x - (-2x) \cdot 3 \\
&= -2x^3 - 10x^2 + 6x
\end{aligned}$

c) $-\dfrac{1}{2}x^2(2x^3 - 4x + 3)$

$\begin{aligned}
&= \left(-\dfrac{1}{2}x^2\right) \cdot 2x^3 - \left(-\dfrac{1}{2}x^2\right) \cdot 4x + \left(-\dfrac{1}{2}x^2\right) \cdot 3 \\
&= -x^5 + 2x^3 - \dfrac{3}{2}x^2
\end{aligned}$

Bài 2:

a) $(2x - 1)(x^2 + 5 - 4)$

$\begin{aligned}
&= (2x - 1)(x^2 + 1) \\
&= 2x \cdot x^2 + 2x \cdot 1 - 1 \cdot x^2 - 1 \cdot 1 \\
&= 2x^3 + 2x - x^2 - 1 \\
&= 2x^3 - x^2 + 2x - 1
\end{aligned}$

b) $7x(x - 4) - (7x + 3)(2x^2 - x + 4)$

$\begin{aligned}
&= (7x^2 - 28x) - (7x \cdot 2x^2 - 7x \cdot x + 7x \cdot 4 + 3 \cdot 2x^2 - 3 \cdot x + 3 \cdot 4) \\
&= 7x^2 - 28x - (14x^3 - 7x^2 + 28x + 6x^2 - 3x + 12) \\
&= 7x^2 - 28x - (14x^3 - x^2 + 25x + 12) \\
&= 7x^2 - 28x - 14x^3 + x^2 - 25x - 12 \\
&= -14x^3 + 8x^2 - 53x - 12
\end{aligned}$

c) $-(5x - 4)(2x + 3)$

$\begin{aligned}
&= -(5x \cdot 2x + 5x \cdot 3 - 4 \cdot 2x - 4 \cdot 3) \\
&= -(10x^2 + 15x - 8x - 12) \\
&= -(10x^2 + 7x - 12) \\
&= -10x^2 - 7x + 12
\end{aligned}$

d) $(6x^5y^2 - 9x^4y^3 + 15x^3y^4) : 3x^3y^2$

$\begin{aligned}
&= \dfrac{6x^5y^2}{3x^3y^2} - \dfrac{9x^4y^3}{3x^3y^2} + \dfrac{15x^3y^4}{3x^3y^2} \\
&= 2x^2 - 3xy + 5y^2
\end{aligned}$

e) $(2x^3 - 21x^2 + 67x - 60) : (x - 5)$

$\begin{aligned}
&= (2x^3 - 10x^2 - 11x^2 + 55x + 12x - 60) : (x - 5) \\
&= [2x^2(x - 5) - 11x(x - 5) + 12(x - 5)] : (x - 5) \\
&= (x - 5)(2x^2 - 11x + 12) : (x - 5) \\
&= 2x^2 - 11x + 12
\end{aligned}$