\(a.=x^2-2x\cdot2y+\left(2y\right)^2=\left(x-2y\right)^2.\)

\(b.=\left(3xy\right)^2-2\cdot3xy\cdot1+1^2=\left(3xy-1\right)^2.\)

\(c.=x^3-3x^2+3x-1-1=x^3-3x^2+3x-2\)

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

\(d.x^3-3x^2+3x-1=\left(x-1\right)^3.\)

あなたはとても正しいです

\(ax+bx+cx+a+b+c\)

\(=a\left(x+1\right)+b\left(x+1\right)+c\left(x+1\right)\)

\(=\left(1+x\right)\left(a+b\right)\left(1+x\right)c=\left(1+x\right)\left(a+b+c\right)\)

\(a.\left(x^2-3x+1\right)\left(-2x-5\right)=\left(-2x-5\right)\left(x^2-3x+1\right)\)

\(=-2x\left(x^2-3x+1\right)-5\left(x^2-3x+1\right)\)

\(=-2x^3+6x^2-2x-5x^2+15x-5\)

\(=-2x^3+x^2+13x-5.\)

\(b.\left(3x^4-8x^3-10x^2+8x-5\right)\left(3x^2-2x+1\right)\)

\(=\left(3x^2-2x+1\right)\left(3x^4-8x^3-10x^2+8x-5\right)\)

\(=9x^6-30x^5-11x^4+36x^3-41x^2+18x-5.\)

\(c.12x^5y+24x^2y^2-36x^3y^2\)

\(=12x^2y\cdot x^3+12x^2y\cdot2-12x^2y\cdot3xy=12x^2y\left(x^3+2-3xy\right).\)

\(d.4x^2-3x-6y+8xy=\left(4x^2-3x\right)+\left(-6y+8xy\right)\)

\(=x\left(4x-3\right)+2y\left(4x-3\right)=\left(4x-3\right)\left(x+2y\right).\)

\(e.x^2-5x+4=\left(x^2-x\right)+\left(-4x+4\right)=x\left(x-1\right)-4\left(x-1\right)\)

\(=\left(x-1\right)\left(x-4\right).\)

nó bị lỗi rồi 

ok

\(a.\left(\dfrac{1}{2}\right)^5\cdot\left(\dfrac{1}{4}\right)^2=\left(\dfrac{1}{2}\right)^5\cdot\left(\dfrac{1}{2}\right)^4=\left(\dfrac{1}{2}\right)^9=\dfrac{1}{512}.\)

\(b.\left(\dfrac{5}{4}\right)^2:\left(-\dfrac{35}{24}\right)^2=\dfrac{25}{16}:\dfrac{1225}{567}=\dfrac{25}{16}\cdot\dfrac{567}{1225}=\dfrac{36}{49}.\)

\(c.\left(-\dfrac{1}{2}\right)^2\cdot\left(\dfrac{2}{5}\right)^2=\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}.\)

\(d.25\cdot5^{-1}\cdot5^0=25\cdot\left(5^{-1}\cdot5^0\right)=25\cdot5^{-1}=25\cdot\dfrac{1}{5}=5.\)