Proszę o obliczenia wszystkie!!! A) (1/3) do potęgi 3 + 3/7 x 2,1 -2/3 x (2,25:2,5) B) 1 5/7 x (0,28.... Question from @malutkaaqvq

April 2022 1 4 Report

Proszę o obliczenia wszystkie!!!
A) (1/3) do potęgi 3 + 3/7 x 2,1 -2/3 x (2,25:2,5)
B) 1 5/7 x (0,28:0,4) -(0,1) do potęgi 2 +0,9: 2 1/4
Oraz proszę o pilną odpowiedź!!

SpongeBob380

$a) \: \: ( \frac{ 1 }{ 3 } ) ^ { 3 } + \frac{ 3 }{ 7 } \times 2.1- \frac{ 2 }{ 3 } \times (2.25 \div 2.5) = \frac{1}{27}+\frac{3}{7}\times 2.1-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{1}{27}+\frac{3}{7}\times \left(\frac{21}{10}\right)-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{1}{27}+\frac{63}{70}-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{1}{27}+\frac{9}{10}-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{10}{270}+\frac{243}{270}-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{253}{270}-\frac{2}{3}\times \left(\frac{2.25}{2.5}\right) = \frac{253}{270}-\frac{2}{3}\times \left(\frac{225}{250}\right) = \frac{253}{270}-\frac{2}{3}\times \left(\frac{9}{10}\right) = \frac{253}{270}-\frac{18}{30} = \frac{253}{270}-\frac{3}{5} = \frac{253}{270}-\frac{162}{270} = \frac{91}{270}$

$b) \: \: 1 \frac{ 5 }{ 7 } \times (0.28 \div 0.4)-(0.1) ^ { 2 } +0.9 \div 2 \frac{ 1 }{ 4 } = \frac{12}{7}\times \left(\frac{0.28}{0.4}\right)-0.1^{2}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{12}{7}\times \left(\frac{28}{40}\right)-0.1^{2}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{12}{7}\times \left(\frac{7}{10}\right)-0.1^{2}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{12}{10}-0.1^{2}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{6}{5}-0.1^{2}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{6}{5}-0.01+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{6}{5}-\frac{1}{100}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{119}{100}+\frac{0.9}{\frac{2\times 4+1}{4}} = \frac{119}{100}+\frac{0.9\times 4}{2\times 4+1} = \frac{119}{100}+\frac{3.6}{2\times 4+1} = \frac{119}{100}+\frac{3.6}{8+1} = \frac{119}{100}+\frac{3.6}{9} = \frac{119}{100}+\frac{2}{5} = \frac{119}{100}+\frac{40}{100} = \frac{159}{100} = 1 \frac{59}{100}$