Question

Giúp vs

1

a. (x-7).(x+7)=0

b.(x-5).(x-9)=0

c.(x-5).(x mũ 2-9)=0

d.(xmu2-7).(xmu2-51) <0

Thanks mấy man đã TL

Answer 1

a,\(\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy...

b,\(\left(x-5\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

Vậy...

c,\(\left(x-5\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy...

Câu d bạn viết lại đề nhé

Answer 2

Theo mink làm như zậy nè

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

Ta xét 2 TH

TH1 nếu x-7 =0 thì x=7

\(\Rightarrow\)x+7 ta tìm đc x là x\(\in\)R

TH2 nêu x+7=0 thì x=-7

\(\Rightarrow\)x-7 ta tìm được x là \(x\in R\)

Câu b,c,d tương tự nhé

Answer 3

a, (x-7).(x+7) = 0
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0+7\\x=0-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

b,(x-5).(x-9)=0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+5\\x=0+9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)