a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)
a/ \(x^2-25=0\)
\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
b/ \(x\left(x+7\right)+x+7=0\)
\(x\left(x+7\right)+\left(x+7\right)=0\)
\(\left(x+7\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
a) x^2 - 25 = 0 <=> (x + 5)(x - 5) = 0
<=> x = -5 hoặc x = 5
b) x(x + 7) + x + 7 = 0 <=> (x + 7)(x + 1) = 0
<=> x = -7 hoặc x = -1
\(a,\Leftrightarrow x^2=25\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\\ b,\Leftrightarrow x\left(x+7\right)+\left(x+7\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-7\end{matrix}\right.\)
a)x2-25=0
<=>x2-52=0
<=>(x-5)(x+5)=0
<=>x-5=0 hoặc x+5=0
Th1:x-5=0 Th2:x+5=0
<=>x=5 <=>x=-5
Vậy x∈{5;-5}
b)x(x+7)+x+7=0
<=>(x+1)(x+7)=0
<=>x+1=0 hoăc x+7=0
Th1:x+1=0 Th2:x+7=0
<=>x=-1 <=>x=-7
Vậy x∈{-1;-7}
