=>x(x-15)=0

=>x=0 hoặc x-15=0

=>x=0 hoặc x=15

\(x^2-15x=0\)

\(\Leftrightarrow x\left(x-15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=15\end{matrix}\right.\)

Vậy \(S=\left\{0;15\right\}\)

15 \(\times\) ( 2\(x\) - 16) - (6\(x^2\) + 15\(x\)): 3\(x\) = 20

15 \(\times\) (2\(x\) - 16) - 3\(x\)( 2\(x\) +  5):3\(x\) = 20

30\(x\) - 240 - (2\(x\) + 5) = 20

30\(x\) - 240 - 2\(x\) - 5 = 20

28\(x\) - 245 = 20

28\(x\)           = 20 + 245

28\(x\)          = 265

     \(x\)         = 265:28

15(2x-16)-(6\(x^2\)+15x):3x=20

=>30x-240-2x-5=20

=>28x=265

=>x=\(\dfrac{265}{28}\)

3x² +15x = 0

3x( x + 5 ) = 0

TH1: 3x = 0

           x = 0 : 3

           x = 0

TH2: x + 5 = 0

        x        = 0 - 5

        x        = -5

Vậy x ∈ { 0; -5 }

a) Ta có: \(7x\left(x-20\right)-x+20=0\)

\(\Leftrightarrow\left(x-20\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=\dfrac{1}{7}\end{matrix}\right.\)

b) Ta có: \(x^3-15x=0\)

\(\Leftrightarrow x\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)

nha nha\

\(\frac{1}{6x}+\frac{1}{10x}-\frac{4}{15x}+1=0\)

\(\Rightarrow\frac{1}{x}\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(\Rightarrow\frac{1}{x}.0+1=0\)

\(\Rightarrow1=0\)( vô lý )

=> không có giá trị x thỏa mãn đề bài 

Study well 

\(\frac{1}{6x}+\frac{1}{10x}-\frac{4}{15x}+1=0\)

\(\Rightarrow\frac{1}{12}\times\frac{2}{x}+\frac{1}{20}\times\frac{2}{x}-\frac{2}{15}\times\frac{2}{x}=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{12}+\frac{1}{20}-\frac{2}{15}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}-\frac{2}{3\times5}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}-\frac{1}{3}-\frac{1}{5}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times0=-1\)

\(\Rightarrow x\in\varnothing\)