a)Ta có: \(x^3-3x^2+2=x^3-x^2-2x^2+2\)

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

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

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

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

\(\left(x-3\right)-\left(x-3\right)\left(x^2+3x+9\right)\left(9x+1\right)=15\)

\(\Leftrightarrow\left(x-3\right)-\left(x^3-27\right)\left(9x+1\right)=15\)

( còn lại tự túc )

a)Ta có:

5.4.7 +516 =5.4.7+4.129=4.(5.7+129) chia hết cho 4

=>5.4.7+516 là hợp số

b)ta có:

25.2-9.5 =5.5.2-9.5=5.(5.2-9) chia hết cho 5

=>25.2-9.5 là hợp số

 31+(-7)-40+|-28|

=>31+(-7)-40+28

   =24+(-7)-40+28

   =17-40+28

   =-23+28

   =5

\(4x^2-12x+9=\left(5-x\right)^2\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot3+3^2=\left(5-x\right)^2\)

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

\(\Leftrightarrow\left(2x-3\right)^2-\left(5-x\right)^2=0\)

\(\Leftrightarrow\left(2x-3+5-x\right)\left(2x-3-5+x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\3x-8=0\Leftrightarrow x=\dfrac{8}{3}\end{matrix}\right.\)

Ta có:

\(A=5x^2-\left[4x^2-3x\left(x-20\right)\right]\)

\(\Leftrightarrow A=5x^2-\left[4x^2-\left(3x^2-60x\right)\right]\)

\(\Leftrightarrow A=5x^2-\left(4x^2-3x^2+60x\right)\)

\(\Leftrightarrow A=5x^2-\left(x^2+60x\right)\)

\(\Leftrightarrow A=5x^2-x^2-60x\)

\(\Leftrightarrow A=4x^2-60x\)

Thay \(x=-\dfrac{1}{2}\) vào biểu thức \(A=4x^2-60x\), ta được:

\(A=4\cdot\left(-\dfrac{1}{2}\right)^2-60\cdot\dfrac{-1}{2}=4\cdot\dfrac{1}{4}+30=31\)

\(\dfrac{x-1}{2y}\cdot\dfrac{x-1}{2y}=\dfrac{\left(x-1\right)^2}{4y^2}=\dfrac{x^2-2x+1}{4y^2}\)

a)\(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)+4\left(x^{n+1}+2y^{n-1}\right)\)

\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)

\(=4x^{n+1}-8y^{n-1}\) \(\left(=4\left(x^{n+1}-2y^{n-1}\right)\right)\)