\(a,x\left(x-5\right)-5+x=0\)
\(\Leftrightarrow x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
a) \(x\left(x-5\right)=5-x\)
\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{-1;5\right\}\)
b) \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow6\left(x-1\right)=0\) \(\Leftrightarrow x=1\)
Vậy \(x=1\)
c) \(\left(x+2\right)^2=x+2\) \(\left(ĐK:x\ge-2\right)\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\) (Thỏa mãn)
Vậy \(x\in\left\{-2;-1\right\}\)
d) \(x^3-2x=0\)
\(\Leftrightarrow x\left(x^2-2\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm\sqrt{2}\right\}\)
a) x(x-5)=5-x
=> x(x-5) + (5-x)=0
=> x(x-5) - (x-5)=0
=> (x-5)(x-1)=0
=>\(\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) (2x+3)(x-1)+(2x-3)(1-x)=0
=>(2x+3)(x-1)+[-(2x-3)](x-1)=0
=>(x-1)[-(2x-3)+(2x+3)]=0
=>(x-1)[-2x+3+2x+3]=0
=>(x-1)6=0
=>x-1=0
=>x=1
c) (x+2)2=x+2
=> (x+2)(x+2) - (x+2)=0
=> (x+2)[(x+2)-1]=0
=>(x+2)[x+1]=0
=>\(\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)
d)x3-2x=0
=>x(x2-2)=0
=>\(\left[{}\begin{matrix}x=0\\x^2-2=0\Rightarrow x^2=2\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
