a,
\(y^2-x^2+10x-25\)
\(=y^2-\left(x^2-10x+25\right)\)
\(=y^2-\left(x-5\right)^2\)
\(=\left(y+x-5\right)\left(y-x+5\right)\)
a) \(y^2-x^2+10x-25=y^2-\left(x^2-10x+25\right)=y^2-\left(x^2-2.x.5+5^2\right)\)
\(=y^2-\left(x-5\right)^2=\left(y-x+5\right).\left(y+x-5\right)\)
b) \(\left(3x+1\right)^2=3x+1\Rightarrow\left(3x-1\right)^2-\left(3x+1\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(3x+1-1\right)=0\Rightarrow\left(3x+1\right).3x=0\)
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=0\end{cases}}}\)
Ta có
\(\left(3x+1\right)^2=3x+1\)
\(\left(3x+1\right)^2-\left(3x+1\right)=0\)
\(\left(3x+1\right)\left(3x+1-1\right)=0\)
\(x\left(3x+1\right)=0\)\(\Rightarrow\orbr{\begin{cases}x=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}}\)
\(y^2-x^2+10x-25\)
\(=y^2-\left(x^2-10x+25\right)\)
\(=y^2-\left(x-5\right)^2\)
\(=\left(y-x+5\right)\left(y+x-5\right)\)
\(\left(3x+1\right)^2=3x+1\)
\(\left(3x+1\right)^2-\left(3x+1\right)=0\)
\(\left(3x+1\right)3x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=0\end{cases}}\)
Vậy ....
