Question

tim cac gia tri cua x de bieu thuc sau co gia tri bang 0:

P(x)=2x-7+(x-14)

Q(x)=x^2-64

Answer 1

a/ Ta có :

\(P\left(x\right)=2x-7+\left(x-14\right)\)

Mà \(P\left(x\right)=0\)

\(\Leftrightarrow2x-7+\left(x-14\right)=0\)

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

\(\Leftrightarrow\left(2x-x\right)+\left(14-7\right)=0\)

\(\Leftrightarrow x+7=0\)

\(\Leftrightarrow x=-7\)

Vậy ...

b/ Ta có :

\(Q\left(x\right)=x^2-64\)

Mà \(Q\left(x\right)=0\)

\(\Leftrightarrow x^2-64=0\)

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

Vậy ...