Question

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7/5-[7/3-{1/1+(5/2+4/5-1/2)}]​

Answer 1

Step-by-step explanation:

$\frac{ 7 }{ 5 } ( \frac{ 7 }{ 3 } -( \frac{ 1 }{ 1 } +( \frac{ 5 }{ 2 } + \frac{ 4 }{ 5 } - \frac{ 1 }{ 2 } )))$

$\frac{7}{5}\left(\frac{7}{3}-\left(1+\frac{5}{2}+\frac{4}{5}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{2}{2}+\frac{5}{2}+\frac{4}{5}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{2+5}{2}+\frac{4}{5}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{7}{2}+\frac{4}{5}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{35}{10}+\frac{8}{10}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{35+8}{10}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\left(\frac{43}{10}-\frac{1}{2}\right)\right)$

$\frac{7}{5}\left(\frac{7}{3}-\frac{43-5}{10}\right)$

$\frac{7}{5}\left(\frac{7}{3}-\frac{38}{10}\right)$

$\frac{7}{5}\left(\frac{7}{3}-\frac{19}{5}\right)$

$\frac{7}{5}\left(\frac{35}{15}-\frac{57}{15}\right)$

$\frac{7}{5}\times \left(\frac{35-57}{15}\right)$

$\frac{7}{5}\left(-\frac{22}{15}\right)$

$\frac{7\left(-22\right)}{5\times 15}$

$\frac{-154}{75}$

$-\frac{154}{75}$

hope it helps

Answer 2

★ How to do :-

Here we are given an expression with fractions given inside brackets. We are asked to simplify the fractions by adding and subtracting them. This can be implemented by using the correct mathematical operations. Here, we use the concept of BODMAS which says that first, we should solve the brackets. Then, we should divide the fractions, next the multiplication of numbers, then comes the addition and finally the subtraction. The name of the rule involves the first letter of every operations on numbers. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \dfrac{7}{5} - \bigg[ \dfrac{7}{3} - \bigg\{ \dfrac{1}{1} + \bigg( \dfrac{5}{2} + \dfrac{4}{5} - \dfrac{1}{2} \bigg) \bigg\} \bigg]}$

First solve the round bracket.

${\tt \leadsto \dfrac{5}{2} + \dfrac{4}{5} - \dfrac{1}{2}}$

LCM of 5 and 2 is 10.

${\tt \leadsto \dfrac{5 \times 5}{2 \times 5} + \dfrac{4 \times 2}{5\times 2} - \dfrac{1 \times 5}{2 \times 5}}$

Multiply the numbers on numerators and denominator.

${\tt \leadsto \dfrac{25}{10} + \dfrac{8}{10} - \dfrac{5}{10}}$

Add the fractions first.

$\leadsto \tt{\dfrac{25+8-5}{10}=\dfrac{33-5}{10}}$

Subtract them now.

${\tt \leadsto \dfrac{28}{10}}$

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${\tt \leadsto \dfrac{7}{5} - \bigg[ \dfrac{7}{3} - \bigg\{ \dfrac{1}{1} + \dfrac{28}{10} \bigg\} \bigg]}$

Now, solve the second bracket.

${\tt \leadsto \dfrac{1}{1} + \dfrac{28}{10}}$

LCM of 1 and 10 is 10.

${\tt \leadsto \dfrac{1 \times 10}{1 \times 10} + \dfrac{28}{10}}$

Multiply the numerator and denominator of first fraction.

${\tt \leadsto \dfrac{10}{10} + \dfrac{28}{10}}$

Add the fractions now.

${\tt \leadsto \dfrac{10 + 28}{10} = \dfrac{38}{10}}$

$\:$

${\tt \leadsto \dfrac{7}{5} - \bigg[ \dfrac{7}{3} - \dfrac{38}{10} \bigg]}$

Now, solve the square brackets.

${\tt \leadsto \dfrac{7}{3} - \dfrac{38}{10}}$

LCM of 3 and 10 is 30.

${\tt \leadsto \dfrac{7 \times 10}{3 \times 10} - \dfrac{38 \times 3}{10 \times 3}}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \dfrac{70}{30} - \dfrac{114}{30}}$

Subtract them now.

${\tt \leadsto \dfrac{70 - 114}{30} = \dfrac{(-44)}{30}}$

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${\tt \leadsto \dfrac{7}{5} - \dfrac{(-44)}{30}}$

LCM of 5 and 30 is 30.

${\tt \leadsto \dfrac{7 \times 6}{5 \times 6} - \dfrac{(-44)}{30}}$

Multiply the numerator and denominator of first fraction.

${\tt \leadsto \dfrac{42}{30} - \dfrac{(-44)}{30}}$

Subtract the values now.

${\tt \leadsto \dfrac{42 - (-44)}{5} = \dfrac{86}{30}}$

Write it as a mixed fraction.

${\tt \leadsto \dfrac{86}{30} = 2 \dfrac{26}{30}}$

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$\red{\underline{\boxed{\bf So, \: the \: answer \: when \: simplified \: is \: 2 \dfrac{26}{30}}}}$