Question

Tìm x, biết:

\(5x. (x - 7)- 40.(x - 7) = 0 \)

\(x. ( x + 5 )- 7 ( x + 5 ) = 0\)

Answer 1

5x( x - 7 ) - 40( x - 7 ) = 0

<=> ( x - 7 )( 5x - 40 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\5x-40=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

x( x + 5 ) - 7( x + 5 ) = 0

<=> ( x + 5 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)

Answer 2

Ta có: \(5x.\left(x-7\right)-40.\left(x-7\right)=0\)

=> \(\orbr{\begin{cases}5x-40=0\\x-7=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

Ta có: \(x.\left(x+5\right)-7.\left(x+5\right)=0\)

=> \(\orbr{\begin{cases}x-7=0\\x+5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)

Mình làm hơi tắt xíu :v ~~~

Answer 3

\(5x.\left(x-7\right)-40.\left(x-7\right)=0\)

\(\left(5x-40\right).\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-40=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=40\\x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=7\end{cases}}}\)

\(x.\left(x+5\right)-7.\left(x+5\right)=0\)

\(\left(x-7\right).\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0+7\\x=0-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)

Answer 4

a) \(5x.\left(x-7\right)-40.\left(x-7\right)=0\)

\(\left(x-7\right).\left(5x-40\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\5x-40=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

b) \(x.\left(x+5\right)-7.\left(x+5\right)=0\)

\(\left(x+5\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-7=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)