Lời giải:
a. $(x-1)(5x+3)=(3x-8)(x-1)$
$\Leftrightarrow (x-1)(5x+3)-(x-1)(3x-8)=0$
$\Leftrightarrow (x-1)(5x+3-3x+8)=0$
$\Leftrightarrow (x-1)(2x+11)=0$
$\Leftrightarrow x-1=0$ hoặc $2x+11=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{-11}{2}$
b.
$2x(25x+15)-35(5x+3)=0$
$\Leftrightarrow 10x(5x+3)-35(5x+3)=0$
$\Leftrightarrow (5x+3)(10x-35)=0$
$\Leftrightarrow 5x+3=0$ hoặc $10x-35=0$
$\Leftrightarrow x=\frac{-3}{5}$ hoặc $x=\frac{35}{10}$
c.
$(2-3x)(x+11)=(3x-2)(2-5x)$
$\Leftrightarrow (2-3x)(x+11)-(3x-2)(2-5x)=0$
$\Leftrightarrow (2-3x)(x+11)-(2-3x)(5x-2)=0$
$\Leftrightarrow (2-3x)(x+11-5x+2)=0$
$\Leftrightarrow (2-3x)(13-4x)=0$
$\Leftrightarrow 2-3x=0$ hoặc $13-4x=0$
$\Leftrightarrow x=\frac{2}{3}$ hoặc $x=\frac{13}{4}$
\(a,\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{2}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{11}{2}\right\}\)
\(b,\Leftrightarrow3x.5\left(5x+3\right)-35\left(5x+3\right)=0\left(tách\right)\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15x-35=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
\(Vậy\cdot S=\left\{\dfrac{7}{3};-\dfrac{3}{5}\right\}\)
d.
$(2x^2+1)(4x-3)=(2x^2+1)(x-12)$
$\Leftrightarrow (2x^2+1)(4x-3)-(2x^2+1)(x-12)=0$
$\Leftrightarrow (2x^2+1)(4x-3-x+12)=0$
$\Leftrightarrow (2x^2+1)(3x+9)=0$
$\Leftrightarrow 3x+9=0$ (do $2x^2+1>0$ với mọi $x\in\mathbb{R}$)
$\Leftrightarrow x=-3$
e.
$(2x-1)^2+(2-x)(2x-1)=0$
$\Leftrightarrow (2x-1)(2x-1+2-x)=0$
$\Leftrightarrow (2x-1)(x+1)=0$
$\Leftrightarrow 2x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=-1$
f.
$(x+2)(3-4x)=x^2+4x+4=(x+2)^2$
$\Leftrightarrow (x+2)^2-(x+2)(3-4x)=0$
$\Leftrightarrow (x+2)(x+2-3+4x)=0$
$\Leftrightarrow (x+2)(5x-1)=0$
$\Leftrightarrow x=-2$ hoặc $x=\frac{1}{5}$
