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Simplify: 7310+3-256+5-3215+32 - Mathematics

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प्रश्न

Simplify:

`(7sqrt(3))/(sqrt(10) + sqrt(3)) - (2sqrt(5))/(sqrt(6) + sqrt(5)) - (3sqrt(2))/(sqrt(15) + 3sqrt(2))`

सरल रूप दीजिए

उत्तर

`(7sqrt(3))/(sqrt(10) + sqrt(3)) - (2sqrt(5))/(sqrt(6) + sqrt(5)) - (3sqrt(2))/(sqrt(15) + 3sqrt(2))`

Rationalise the denominators:

⇒ `((7sqrt(3))/(sqrt(10) + sqrt(3)) xx (sqrt(10) - sqrt(3))/(sqrt(10) - sqrt(3))) - ((2sqrt(5))/(sqrt(6) + sqrt(3)) xx (sqrt(6) - sqrt(5))/(sqrt(6) - sqrt(5))) - ((3sqrt(2))/(sqrt(15) + 3sqrt(2)) xx (sqrt(15) - 3sqrt(2))/(sqrt(15) - 3sqrt(2)))` 

⇒ `(7sqrt(3)(sqrt(10) - sqrt(3)))/(10 - 3) - (2sqrt(5)(sqrt(6) - sqrt(5)))/(6 - 5) - (3sqrt(2)(sqrt(15) - 3sqrt(2)))/(15 - 8)`   ...[∵ a2 – b2 = (a + b)(a – b)]

⇒ `(7sqrt(3)(sqrt(10) - sqrt(3)))/(7) - (2sqrt(5)(sqrt(6) - sqrt(5)))/(1) - (3sqrt(2)(sqrt(15) - 3sqrt(2)))/(3)`

⇒ `(7sqrt(30) - 21)/7 - (2sqrt(30) - 10)/1 + (3sqrt(30) - 18)/3`

⇒ `(21sqrt(30) - 63 - 42sqrt(30) + 210 + 21sqrt(30) - 126)/21`

⇒ `21/21 = 1`

Hence the answer is 1.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Number Systems - Exercise 1.4 [पृष्ठ १२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 1 Number Systems
Exercise 1.4 | Q 2. | पृष्ठ १२

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