Question

tìm x biết: 

a) x^2 - 7x = 0

b) -3x^2 + 5x = 0

Answer 1

Ta có : x² - 7x = 0

=> x(x - 7) = 0

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

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

Ta có : -3x² + 5x = 0

=> \(x\left(-3x+5\right)=0\)

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

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

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

Answer 2

a) x² - 7x = 0

x ( x - 7 ) = 0

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

Vay x = 0 hoac x = 7

b) -3x² + 5x = 0

x( -3x + 5 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\)

1. x= 0

2. -3x + 5 = 0

-3x = -5

x = -5 : ( - 3 )

x = \(\frac{5}{3}\)

Vay x= 0 hoac x = \(\frac{5}{3}\)