(x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> (x - 1)(5x + 3) - (3x - 8)(x - 1)= 0
<=> (x - 1)(5x + 3 - 3x + 8) = 0
<=> (x - 1)(2x + 11) = 0
\(\Leftrightarrow \begin{bmatrix} x - 1 = 0 & & \\ 2x + 11 = 0 & & \end{bmatrix}\)pn bỏ dấu ngoặc bên phải nha
\(\Leftrightarrow \begin{bmatrix} x = 1 & & \\ x = \frac{-11}{2} & & \end{bmatrix}\)
Vậy ............
\(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Chúc bạn học tốt!
(x-1)(5x+3)=(3x-8)(x-1)
<=>(x-1)(5x+3)-(x-1)(3x-8)=0
<=>(x-1)(5x+3-3x+8)=0
<=>(x-1)(2x+11)=0
=> x-1= hoặc 2x+11=0
=> x=1 x = \(-\dfrac{11}{2}\)
Vậy S={1;\(-\dfrac{11}{2}\)}
\(a)(x−1)(5x+3)=(3x−8)(x−1)\)
\(\Leftrightarrow (x−1)(2x+3−3x+8)=0\)
\(\Leftrightarrow (x−1)(11−x)=0 \)
\(\Leftrightarrow x−1=0x−1=0 hoặc 11−x=0\)
\(\Leftrightarrow x=1x=1 hoặc x=11\)
Giups mk vs thank cacs bn 
