a: \(x^3-25x=0\)
=>\(x\cdot x^2-x\cdot25=0\)
=>\(x\left(x^2-25\right)=0\)
=>\(x\left(x-5\right)\left(x+5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
b: \(x^4+4=5x^2\)
=>\(x^4-5x^2+4=0\)
=>\(x^4-x^2-4x^2+4=0\)
=>\(\left(x^4-x^2\right)-\left(4x^2-4\right)=0\)
=>\(x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
=>\(\left(x^2-1\right)\left(x^2-4\right)=0\)
=>\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\\x=-2\end{matrix}\right.\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-2x\right)=0\)
=>\(x\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
d: \(4\left(x-2\right)^2=25\left(1-2x\right)^2\)
=>\(25\left(2x-1\right)^2=4\left(x-2\right)^2\)
=>\(\left[5\left(2x-1\right)\right]^2-\left[2\left(x-2\right)\right]^2=0\)
=>\(\left(10x-5\right)^2-\left(2x-4\right)^2=0\)
=>\(\left(10x-5-2x+4\right)\cdot\left(10x-5+2x-4\right)=0\)
=>\(\left(8x-1\right)\left(12x-9\right)=0\)
=>\(\left[{}\begin{matrix}8x-1=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{4}\end{matrix}\right.\)
e: \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
=>\(6x^2-3x-10x+5-\left(6x^2-x+12x-2\right)=0\)
=>\(6x^2-13x+5-6x^2-11x+2=0\)
=>-24x+7=0
=>-24x=-7
=>\(x=\dfrac{7}{24}\)
f: \(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=5\)
=>\(\left(3x\right)^2-2^2-\left(9x^2-6x+1\right)=5\)
=>\(9x^2-4-9x^2+6x-1=5\)
=>6x-5=5
=>6x=10
=>x=\(\dfrac{10}{6}=\dfrac{5}{3}\)
