Math, asked by rocktwosms7367, 1 year ago

Verify [3/4+-2/5]+-7/10=3/4+[-2/5+-7/10]

Answers

Answered by gayatrikumari99sl
6

Answer:

Proved  [\frac{3}{4} +\frac{-2}{5} ]+\frac{-7}{10}  =\frac{3}{4} +[\frac{-2}{5} +\frac{-7}{10}]

Step-by-step explanation:

For the given question we need to proof that

[\frac{3}{4} +\frac{-2}{5} ]+\frac{-7}{10}  =\frac{3}{4} +[\frac{-2}{5} +\frac{-7}{10}] ................(i)

proof :

Solve LHS part

Therefore,  we have  

[\frac{3}{4} +\frac{-2}{5} ]+\frac{-7}{10}

First solve this part  [\frac{3}{4} +\frac{-2}{5} ] , so find the LCM of 4 and 5 .

Step1:

⇒     [\frac{15-8}{20}]+\frac{-7}{10}            (where 20 is the LCM of 4 and 5 )

[\frac{7}{20} ]+\frac{-7}{10}

Again take the LCM of 20 and 10 which is 20 ,

Now   [\frac{7-14}{20} ] = [\frac{-7}{20} ].

So, LHS = [\frac{-7}{20} ]  .

Solve RHS:

Step2:

Similarly from step1

Here we have ,

\frac{3}{4} +[\frac{-2}{5} +\frac{-7}{10}]   = \frac{3}{4} +[\frac{-4-7}{10}]    (where LCM of 5 and 10 is 10 )

⇒   \frac{3}{4} +[\frac{-11}{10}]

Again find LCM of 4 and 10 which is 20 .

Therefore ,

[\frac{15-22}{20}] = \frac{-7}{20}.

Final answer:

Hence , here we proved that LHS = RHS which is  [\frac{3}{4} +\frac{-2}{5} ]+\frac{-7}{10}  =\frac{3}{4} +[\frac{-2}{5} +\frac{-7}{10}].

Answered by qwsuccess
2

Given: Two expressions [\frac{3}{4} + (\frac{-2}{5})] + (\frac{-7}{10}) and \frac{3}{4} + [(\frac{-2}{5}) + (\frac{-7}{10})]

To verify: [\frac{3}{4} + (\frac{-2}{5})] + (\frac{-7}{10}) =  \frac{3}{4} + [(\frac{-2}{5}) + (\frac{-7}{10})]

Solution:

Taking L.H.S.

[\frac{3}{4} + (\frac{-2}{5})] + (\frac{-7}{10})

L.C.M of 4 and 5 = 20

[\frac{5*3 \ + \ 4(-2)}{20}] + (\frac{-7}{10})

[\frac{15 - 8}{20}] + (\frac{-7}{10}) = (\frac{7}{20}) + (\frac{-7}{10})

L.C.M of 10 and 20 = 20

[\frac{7 \ + \ 2(-7)}{20} ] = \frac{7 \ - \ 14}{20} = \frac{-7}{20}

Taking R.H.S.

⇒  \frac{3}{4} + [(\frac{-2}{5}) + (\frac{-7}{10})]

L.C.M of 5 and 10 = 10

\frac{3}{4} + [\frac{2(-2) \ + \ (-7)}{10}]

\frac{3}{4} + [\frac{-4 \ - 7}{10}  ] = \frac{3}{4} + (\frac{-11}{10})

L.C.M of 4 and 10 = 20

[\frac{5*3 \ + \ 2(-11)}{20} ] = \frac{15 - 22}{20} = \frac{-7}{20}

∴ L.H.S = R.H.S = \frac{-7}{20}

Hence, verified that   [\frac{3}{4} + (\frac{-2}{5})] + (\frac{-7}{10}) =  \frac{3}{4} + [(\frac{-2}{5}) + (\frac{-7}{10})]

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