Giải thích các bước giải:

$\begin{aligned}

&\text{a) Ta có:}\\
& VP = \dfrac{14x^2}{-4x} \\
& = \dfrac{2x \cdot 7x}{2x \cdot (-2)} \\
& = \dfrac{7x}{-2} \\
& = \dfrac{7x \cdot y}{-2 \cdot y} \\
& = \dfrac{7xy}{-2y} \\
& = VT \\
\\
&\text{b) Ta có:}\\& VP = \dfrac{3x(x+y)^2}{9x^2(x+y)} \\
& = \dfrac{3x(x+y) \cdot (x+y)}{3x(x+y) \cdot 3x} \\
& = \dfrac{x+y}{3x} \\
& = VT \\
\\
&\text{c) Ta có:}\\& VP = \dfrac{x^2+4x+3}{x^2+6x+9} \\
& = \dfrac{x^2+x+3x+3}{(x+3)^2} \\
& = \dfrac{x(x+1) + 3(x+1)}{(x+3)^2} \\
& = \dfrac{(x+1)(x+3)}{(x+3)(x+3)} \\
& = \dfrac{x+1}{x+3} \\
& = VT \\
\\
&\text{d) Ta có:}\\& VP = \dfrac{7x^3y^4}{35xy} \\
& = \dfrac{7xy \cdot x^2y^3}{7xy \cdot 5} \\
& = \dfrac{x^2y^3}{5} \\
& = VT \\
\\
&\text{e) Ta có:}\\& VT = \dfrac{x^2(x+2)}{x(x+2)^2} \\
& = \dfrac{x \cdot x(x+2)}{(x+2) \cdot x(x+2)} \\
& = \dfrac{x}{x+2} \\
& = VP \\
\\
&\text{f) Ta có:}\\& VT = \dfrac{x^3-4x}{10-5x} \\
& = \dfrac{x(x^2-4)}{-5(x-2)} \\
& = \dfrac{x(x-2)(x+2)}{-5(x-2)} \\
& = \dfrac{x(x+2)}{-5} \\
& = \dfrac{x^2+2x}{-5} \\
& = -\dfrac{x^2+2x}{5} \\
& = \dfrac{-x^2-2x}{5} \\
& = VP
\end{aligned}$