(73/10 )^2 (21/10 )^2 =(73/10 )^2 (21/10 )^2 =(73/10+21/10 )(73/10 21/10 ) =(94/10 )(52/10 ) =(47/5 )(26/5 )=1222/25=4822/25 ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation by Regrouping Terms

73/10 2-21/10...

Question

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}$

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Solution

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2} = {(\frac{73}{10})}^{2} - {(\frac{21}{10})}^{2} = (\frac{73}{10} + \frac{21}{10}) (\frac{73}{10} - \frac{21}{10}) = (\frac{94}{10}) (\frac{52}{10}) = (\frac{47}{5}) (\frac{26}{5}) = \frac{1222}{25} = 48 \frac{22}{25}$

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Similar questions

{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

Q. Factorise:

(7 \frac{3}{10})^{2} - (2 \frac{1}{10})^{2}

Q. Find

{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

Q. Factorise by using formula:

{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

48 \frac{22}{25}

If true then enter

1

and if false then enter

0

Factorise:

(i) $4 a^{2} - 25$ (ii) $x^{2} - \frac{9}{16}$ (iii) $x^{4} - y^{4}$ (iv) $[7 \frac{3}{10}]^{2} - [2 \frac{1}{10}]^{2}$ (v) $(0.7)^{2} - (0.3)^{2}$ (vi) $(5 a - 2 b)^{2} - (2 a - b)^{2}$

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(7310)2−(2110)2

(7310)2−(2110)2=(7310)2−(2110)2=(7310+2110)(7310−2110)=(9410)(5210)=(475)(265)=122225=482225

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}$

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2} = {(\frac{73}{10})}^{2} - {(\frac{21}{10})}^{2} = (\frac{73}{10} + \frac{21}{10}) (\frac{73}{10} - \frac{21}{10}) = (\frac{94}{10}) (\frac{52}{10}) = (\frac{47}{5}) (\frac{26}{5}) = \frac{1222}{25} = 48 \frac{22}{25}$