`\color{cyan}\text{Wanna}`

`a)(0,3-x)(0,6x-8)=0`

\(\left[ \begin{array}{l}0,3-x=0\\0,6x-8=0\end{array} \right.\) 

\(\left[ \begin{array}{l}x=0,3\\0,6x=8\end{array} \right.\) 

\(\left[ \begin{array}{l}x=0,3\\x=\frac{8}{0,6}=\frac{40}{3}\end{array} \right.\)

`b)(5x+1)(7-6x)=0`

\(\left[ \begin{array}{l}5x+1=0\\7-6x=0\end{array} \right.\) 

\(\left[ \begin{array}{l}5x=-1\\6x=7\end{array} \right.\) 

\(\left[ \begin{array}{l}x=-\frac{1}{5}\\x=\frac{7}{6}\end{array} \right.\) 

`c)(2x-3)^2-9=0`

`(2x-3+3)(2x-3-3)=0`

\(\left[ \begin{array}{l}2x=0\\2x-6=0\end{array} \right.\) 

\(\left[ \begin{array}{l}x=0\\2x=6\end{array} \right.\) 

\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) 

`d)4x+6=5x(x+3)`

`4x+6-5x^2-15x=0`

`-5x^2-11x+6=0`

`Δ=b^2-4ac=(-11)^2-4.(-5).6=241>0`

Vậy phương trình có `2` nghiệm phân biệt `:` 

`x_1=(-b-\sqrt{Δ})/(2a)=(-(-11)-\sqrt{241})/(2.(-5))=(-11+\sqrt{241})/(10)`

`x_2=(-b+\sqrt{Δ})/(2a)=(-(-11)+\sqrt{241})/(2.(-5))=(-11-\sqrt{241})/(10)`

`e)2(6x-9)=4x-10`

`12x-18-4x+10=0`

`8x-8=0`

`8x=8`

`x=1`