\(a,x^2-2xy+y^2=\left(x-y\right)^2\)

\(b,\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

\(c,\left(3x-5y\right)^3=27x^3-135x^2y+225xy^2-125y^3\)\(d,8-125a^3=\left(2-5a\right)\left(4+10a+25a^2\right)\)

\(e,\left(9x+y\right)\left(9x-y\right)=81x^2-y^2\)

\(f,\left(x+3\right)^3=x^3+3x^2+9x+27\)

\(3x^2+4x-4=0\)

\(\Leftrightarrow\)\(3x^2-6x+2x-4=0\)

\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+2=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=2\end{matrix}\right.\)

\(a,x\left(x+2\right)-3x-6=0\)

\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)\(b,\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)

\(\Leftrightarrow x^3+3x^2+3x+1-3x^2-3x=0\)

\(\Leftrightarrow x^3+1=0\)

\(\Rightarrow x^3=1\Rightarrow x=1\)

\(c,4x^2-25=0\)

\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}2x+5=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

okey! Vì you t sẽ chăm thêm 1 lần nữa!!!^^

\(a.-y^2+\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2-y^2=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\)

\(b,4\left(x-3\right)^2-9\left(x+1\right)^2=\left[2\left(x-3\right)\right]^2-\left[3\left(x+1\right)\right]^2=\left(2x-6\right)^2-\left(3x+3\right)^2=\left(2x-6-3x-3\right)\left(2x-6+3x+3\right)=\left(-x-9\right)\left(5x-3\right)\)\(c,25x^2-20xy+4y^2=\left(5x-2y\right)^2\)

\(d,-9x^2+12xy-4y^2=-\left(3x-2y\right)^2\)

\(e,25x^2-\dfrac{1}{8}x^2y^2=\left(5x-\dfrac{\sqrt{2}}{4}xy\right)\left(5x+\dfrac{\sqrt{2}}{4}xy\right)\)\(f,9x^2+6x+1=\left(3x+1\right)^2\)

\(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)

=\(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{\left(2x-1\right)\left(x-1\right)}{x^3-1}-\dfrac{6\left(x^2+x+1\right)}{x^3-1}\)

\(=\dfrac{4x^2-3x+17+2x^2-2x-x+1-6x^2-6x-6}{x^3-1}\)\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)

\(a,x+y=1\Rightarrow\left(x+y\right)^3=1\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Leftrightarrow x^3+y^3+3xy=1\)

\(b,x-y=1\Rightarrow\left(x-y\right)^3=1\)

\(\Leftrightarrow x^3-3x^2y+3xy^2-y^3=1\)

\(\Leftrightarrow x^3-y^3-3xy\left(x-y\right)=1\)

\(\Leftrightarrow x^3-y^3-3xy=1\)

\(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{7}{2}=0\)

\(\Leftrightarrow2\left(x-\dfrac{3}{2}\right)^2=\dfrac{7}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\sqrt{\dfrac{7}{4}}\\x-\dfrac{3}{2}=-\sqrt{\dfrac{7}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{7}{4}}+\dfrac{3}{2}\\x=-\sqrt{\dfrac{7}{4}}+\dfrac{3}{2}\end{matrix}\right.\)

\(\left(2x^2-1\right)^2-2\left(2x^2-1\right)+1=0\Leftrightarrow\left(2x^2-1-1\right)^2=0\)\(\Leftrightarrow\left(2x^2-2\right)^2=0\)

\(\Rightarrow2\left(x^2-1\right)=0\)

\(\Rightarrow2\left(x+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(a,x^4+2x^3+x^2=\left(x^2+x\right)^2\)

\(b,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)\)\(=\left(x-1\right)\left(x+6\right)\)

\(c,5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\) \(e,x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)\(f,x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)\(h,2x^2+5x-3=0\Leftrightarrow2x^2-6x+x-3=0\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)