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प्रश्न
Simplify the following expressions:
`(sqrt3 + sqrt7)^2`
उत्तर
We know that `(a + b)^2 = a^2 + b^ + 2ab` We will use this property to simplify the expression
`(sqrt3 + sqrt7)^2`
`∴(sqrt3 + sqrt7)^2 = (sqrt3)^2 + (sqrt7)^2 + 2 xx sqrt3 xx sqrt7`
`= sqrt(3 xx 3) + sqrt(7 xx 7) + 2 xx sqrt(3 xx 7)`
`= (3^2)^(1/2) + (7^2)^(1/2) + 2sqrt21`
`= 3^1+ 7^1 + 2sqrt21`
`= 10 + 2sqrt21`
Hence the value of expression is `10 + 2sqrt21`
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संबंधित प्रश्न
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = `c/d`. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Rationalise the denominator of the following:
`1/sqrt7`
Simplify of the following:
`root(3)4 xx root(3)16`
In the following determine rational numbers a and b:
`(sqrt3 - 1)/(sqrt3 + 1) = a - bsqrt3`
In the following determine rational numbers a and b:
`(5 + 3sqrt3)/(7 + 4sqrt3) = a + bsqrt3`
Write the rationalisation factor of \[\sqrt{5} - 2\].
Rationalise the denominator of the following:
`(4sqrt(3) + 5sqrt(2))/(sqrt(48) + sqrt(18))`
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`(sqrt(10) - sqrt(5))/2`
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`sqrt(2)/(2 + sqrt(2)`
Simplify:
`64^(-1/3)[64^(1/3) - 64^(2/3)]`
