a, \(0,\left(27\right)+0,\left(72\right)=0,\left(01\right).27+0,\left(01\right).72\)

\(=\dfrac{1}{99}.\left(27+72\right)=\dfrac{1}{99}.99=1\)

b, \(0,\left(22\right).\dfrac{9}{2}=0,\left(01\right).22.\dfrac{9}{2}=\dfrac{1}{99}.22.\dfrac{9}{2}=1\)

c, \(\left[0,\left(11\right).9\right]^{2003}=\left[0,\left(01\right).11.9\right]^{2003}\)

\(=\left[\dfrac{1}{99}.11.9\right]^{2003}=1^{2003}=1\)

\(0,5=\dfrac{1}{2}\)

\(1,\left(45\right)=1+0,\left(45\right)=1+45.0,\left(01\right)\)

\(=1+45.\dfrac{1}{99}=\dfrac{16}{11}\)

\(0,2\left(13\right)=0,2+0,0\left(13\right)=0,2+0,\left(13\right):10\)

\(=0,2+0,\left(01\right).13:10=0,2+\dfrac{1}{99}.13:10=\dfrac{211}{990}\)

\(0,04\left(3\right)=0,04+0,00\left(3\right)\)

\(=0,04+0,\left(3\right):100=0,04+0,\left(1\right).3:100\)

\(=0,04+\dfrac{1}{9}.3:100=\dfrac{13}{300}\)

a, \(x^3+8=x^3+2x^2-2x^2-4x+4x+8\)

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

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

Câu b hình như sai đề bạn xem lại nha!

Bạn xem lại đề đi

a, \(3x^2-3x+2x^3-2x^2=0\)

\(\Rightarrow3x.\left(x-1\right)+2x^2.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(3x+2x^2\right)=0\)

\(\Rightarrow\left(x-1\right).x.\left(3+2x\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x=0\\3+2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=0\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy......

Câu b tương tự!!