Question

Tìm x:

a,x(3x-5)=0

b,3x^2-27=0

c,(x-5)^2=x-5

d,2(x+7)-x^2-7x=0

e,7x(x-3)+2(3x)=0

Answer 1

\(x\left(3x-5\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\3x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{5}{3}\right\}\)

Answer 2

a) \(x\left(3x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)

b) \(3x^2-27=0\)

\(\Leftrightarrow3x^2=27\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

c) \(\left(x-5\right)^2=x-5\)

\(\Leftrightarrow x^2-10x+25-x+5=0\)

\(\Leftrightarrow x^2-11x+30=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)

d) \(2\left(x+7\right)-x^2-7x=0\)

\(\Leftrightarrow2x+14-x^2-7x=0\)

\(\Leftrightarrow-x^2-5x+14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)

e)\(7x\left(x-3\right)+2.3x=0\)

\(\Leftrightarrow7x^2-21x+6x=0\)

\(\Leftrightarrow7x^2-15x=0\)

\(\Leftrightarrow x\left(7x-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{15}{7}\end{cases}}}\)

#H

Answer 3

\(3x^2-27=0\)

\(3x^2=27\)

\(x^2=9\)

\(x\in\left\{3;-3\right\}\)

Answer 4

\(\left(x-5\right)^2=x-5\)

\(\left(x-5\right)\left(x-5\right)=x-5\)

\(x\left(x-5\right)-5\left(x-5\right)=x-5\)

\(x^2-5x-5x+25=x-5\)

\(x^2-10x+25=x-5\)

\(x^2-10x+25-\left(x-5\right)=0\)

\(x^2-10x+25-x+5=0\)

\(x^2-10x+30-x=0\)

\(x^2-11x+30=0\)

\(\left(x-6\right)\left(x-5\right)=0\)

\(\hept{\begin{cases}x-6=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=6\\x=5\end{cases}}}\)

Vậy \(x\in\left\{6;5\right\}\)