a) (5x + 3)(x² + 4)(x - 4) = 0

<=> 5x + 3 = 0 hoặc x - 4 = 0

<=> x = -3/5 hoặc x = 4

b) (7x - 2x)(2x - 1)(x + 3) = 0

<=> 5x(2x - 1)(x + 3) = 0

<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0

<=> x = 0 hoặc c = 1/2 hoặc x = -3

(5x +3)(x^2+4)(x-4)=0

\(\Rightarrow\hept{\begin{cases}5x+3=0\\x^2+4=0\\x-4=0\end{cases}}\) nhớ là dùng ngoặc vuông nhé vì olm  k cho dùng ngoặc [   cho 3 cái

\(\Leftrightarrow\hept{\begin{cases}5x=3\\x^2=-4\\x=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{5}\\x\in\left\{\varnothing\right\}\\x=4\end{cases}}\)

vậy pt có nghiệm ....

câu b làm tương tự nhé!

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

\(\Leftrightarrow\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}=\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)

\(\Leftrightarrow6x+30-16x+24=18x-3+4x-2\)

\(\Leftrightarrow6x-16x-18x-4x=-2-3-24-30\)

\(\Leftrightarrow-32x=-59\)

\(\Leftrightarrow x=\frac{59}{32}\)

\(\text{GIẢI :}\)

\(A=\left(\frac{x-2}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\left(\frac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\right):\frac{2x-2}{x}\)

\(=\frac{x^2-3x-2x+6-x^2+9}{x\left(x-3\right)}:\frac{2x-2}{x}\)

\(=\frac{-5x+15}{x\left(x-3\right)}\cdot\frac{x}{2x-2}\)

\(=\frac{-5\left(x-3\right)}{x\left(x-3\right)}\cdot\frac{x}{2x-2}=\frac{-5}{2x-2}\).