Question

Answer 1

a: Ta có: \(x^2+10x+25=0\)

\(\Leftrightarrow\left(x+5\right)^2=0\)

\(\Leftrightarrow x+5=0\)

hay x=-5

b: Ta có: \(x^2+\dfrac{2}{5}x+\dfrac{1}{25}=\dfrac{4}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{4}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

d: Ta có: \(x^2-49=0\)

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

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

Answer 2

a,\(x^2+10x+25=0\)

\(\Leftrightarrow\left(x+5\right)^2=0\)

\(\Leftrightarrow x+5=0\)

\(\Leftrightarrow x=-5\)

d,\(x^2-49=0\)

\(\Leftrightarrow x^2-7^2=0\)

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

TH1: x+7=0 =>x=-7

TH2: x-7=0=>x=7

 

 

Answer 3

a) \(x^2+10x+25=0\Rightarrow\left(x+5\right)^2=0\Rightarrow x+5=0\Rightarrow x=-5\)

b) \(x^2+\dfrac{2}{5}x+\dfrac{1}{25}=\dfrac{4}{25}\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{4}{25}\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

d) \(x^2-49=0\Rightarrow\left(x-7\right)\left(x+7\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

e) \(16x^2-1=0\Rightarrow\left(4x-1\right)\left(4x+1\right)=0\Rightarrow\left[{}\begin{matrix}4x-1=0\\4x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)