Question

Tìm x:

x^2(2x+7)=16(2x+7)

Answer 1

Đề:........

<=> x². (2x + 7) - 16. (2x + 7) = 0

<=> (2x + 7). (x² - 16) = 0

<=> (2x+ 7). (x - 4). (x + 4) = 0

=> \(\hept{\begin{cases}2x+7=0\\x-4=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=-7\\x=4\\x=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-7}{2}\\x=4\\x=-4\end{cases}}\)

Vậy...........

Answer 2

\(x^2\left(2x+7\right)=16\left(2x+7\right)\)

\(x^2\left(2x+7\right)-16\left(2x+7\right)=0\)

\(\left(2x+7\right)\left(x^2-16\right)=0\)

\(\left(2x+7\right)\left(x+4\right)\left(x-4\right)=0\)

\(\hept{\begin{cases}x=-\frac{7}{2}\\x=-4\\x=4\end{cases}}\)