\(3.\)
\(a,\)
\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow4x^2-12x+9-x^2-10x-25=0\)
\(\Leftrightarrow3x^2-22x-16=0\)
\(\Leftrightarrow3.\left(x-8\right)\left(x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3=0\left(\text{vô lí}\right)\\x-8=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{8;-\dfrac{2}{3}\right\}\)
\(b,\)
\(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{1;2\right\}\)
\(4.\)
\(a,\)
\(16x^3y+\dfrac{1}{4}yz^3\)
\(=\dfrac{1}{4}y\left(64x^3+z^3\right)\)
\(=\dfrac{1}{4}y\left(4x+z\right)\left(16x^2-4xz+z^2\right)\)
\(b,\)
\(x^{m+4}-x^{m+3}-x-1\)
\(=x^m.x^4-x^m.x^3-x-1\)
\(=x^m.\left(x^4-x^3\right)-x-1\)
\(=x^m.x^3.\left(x+1\right)-\left(x+1\right)\)
\(=\left(x^{m+3}-1\right)\left(x+1\right)\)
3:
a: =>(2x-3-x-5)(2x-3+x+5)=0
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
b: =>x^3-x^2-4(x-1)^2=0
=>x^2(x-1)-4(x-1)^2=0
=>(x-1)(x^2-4x+4)=0
=>x=1 hoặc x=2
