a) \(0,\left(27\right)+0,\left(72\right)=0,\left(100\right)=1\)
b) \(0,\left(22\right)\cdot\dfrac{9}{2}=\dfrac{2}{9}\cdot\dfrac{9}{2}=1\)
c) \(\left[0,\left(11\right)\cdot9\right]^{2003}=\left(\dfrac{1}{9}\cdot9\right)^{2003}=1^{2003}=1\)
a) Ta có :
\(0,\left(27\right)+0,\left(72\right)==\dfrac{27}{99}+\dfrac{72}{99}=\dfrac{99}{99}=1\)
\(\Rightarrow0,\left(27\right)+0,\left(72\right)=1\rightarrowđpcm\)
b) Ta có :
\(0,\left(22\right).\dfrac{9}{2}=\dfrac{2}{9}.\dfrac{9}{2}=\dfrac{18}{18}=1\)
\(\Rightarrow0,22.\dfrac{9}{2}=1\rightarrowđpcm\)
c) Ta có :
\(\left[0,\left(11\right).9\right]^{2003}=\left[\dfrac{1}{9}.9\right]^{2003}=\left[\dfrac{9}{9}\right]^{2003}=1^{2003}=1\)
\(\Rightarrow\left[0,\left(11\right).9\right]^{2003}=1\rightarrowđpcm\)
a) \(0,\left(27\right)+0,\left(72\right)=0,\left(99\right)=1\)
b) \(0,\left(22\right)\cdot\dfrac{9}{2}=\dfrac{2}{9}\cdot\dfrac{9}{2}=1\)
c) \(\left[0,\left(11\right)\cdot9\right]^{2003}=\left(\dfrac{1}{9}\cdot9\right)^{2003}=1^{2003}=1\)
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, \(\left[0,\left(11\right).9\right]^{2003}=\left[0,\left(01\right).11.9\right]^{2003}\)
\(=\left[\dfrac{1}{99}.11.9\right]^{2003}=1^{2003}=1\)
Chúc bạn học tốt!!!
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, \(\left[0,\left(11\right).9\right]^{2003}=\left[0,\left(01\right).11.9\right]^{2003}\)
\(=\left[\dfrac{1}{99}.11.9\right]^{2003}=1^{2003}=1\)
