Question

\(5x\cdot\left(x-\frac{1}{3}\right)=0\)

\(\left(x+\frac{1}{4}\right)\cdot\left(x-\frac{3}{7}\right)=0\)

giup minh voi

 

 

Answer 1

a/\(5x\cdot\left(x-\frac{1}{3}\right)=0\)

Chia làm 2 TH :

TH 1: \(5x=0\Rightarrow x=0\)

TH 2:\(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)

\(\Rightarrow x\in\left\{0;\frac{1}{3}\right\}\)

b/\(\left(x+\frac{1}{4}\right)\cdot\left(x-\frac{3}{7}\right)=0\)

Chia làm 2 Th 

Th1 : \(x+\frac{1}{4}=0\Rightarrow x=-\frac{1}{4}\)

Th2 :\(x-\frac{3}{7}=0\Rightarrow x=\frac{3}{7}\)

\(\Rightarrow x\in\left\{-\frac{1}{4};\frac{3}{7}\right\}\)

Answer 2

1) \(5x\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x=0\\x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

2) \(\left(x+\frac{1}{4}\right)\left(x-\frac{3}{7}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x-\frac{3}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)