Giải:
a) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\)
\(\Leftrightarrow-2\left(2x-1\right)=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c) \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-4\\x=1\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\) (\(x^2\ge0\ne-4\))
Vậy ...
a) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2-25\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
b) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(-2\right)=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c)\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2+4\right)=0\)
Vì \(x^2\ge0\forall x\Rightarrow x^2+4>0\forall x\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(1.x^2-25-\left(x+5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+5\right)-\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+5-1\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
Vậy S={5;-4}
\(2.\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(-2\right)=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
Vậy S=\(\left\{\dfrac{1}{2}\right\}\)