Question

d) 4,1(15).99-13,(12)·100+16,0(21). 10000. pentru cla 8​

Answer 1

Răspuns:

$\red{\boxed{\quad \dfrac{796537}{5} = 159307,4 \quad }}$

Explicație pas cu pas:

$4,1(15) = \dfrac{4115-41}{990}=\dfrac{4074}{990}^{(6}=\dfrac{679}{165}$

$13,(12)=\dfrac{1312-13}{99}=\dfrac{1299}{99}^{(3} =\dfrac{433}{33}$

$16,0(21) = \dfrac{16021-160}{990}=\dfrac{15861}{990}^{(3}=\dfrac{5287}{330}$

$\;$

$4,1(15) \cdot 99 - 13,(12) \cdot 100 + 16,0(21) \cdot 10000=\\\;$

$\dfrac{679}{165} \cdot 99- \dfrac{433}{33} \cdot 100 + \dfrac{5287}{330} \cdot 10000=\\\;$

$\dfrac{679 \cdot 99}{165}^{(33} - \dfrac{433 \cdot 100}{33} + \dfrac{5287\cdot 10000}{330}^{(10}=\\\;$

$\dfrac{679 \cdot 3}{5} - \dfrac{43300}{33} + \dfrac{5287 \cdot 1000}{33}=\\\;$

$\dfrac{2037}{5} - \dfrac{43300}{33} + \dfrac{5287000}{33} = \dfrac{2037}{5} + \dfrac{-43300+5287000}{33} = \\\;$

$\dfrac{2037}{5} +\dfrac{5243700}{33}^{(33} = \dfrac{2037}{5} + 158900 =\\\;$

$\dfrac{2037}{5} + \dfrac{158900 \cdot 5}{5}= \dfrac{2037+794500}{5}=\red{\dfrac{796537}{5}}$