The value of -6 - 2-5- -1 is nbsp-1-5-

The correct option is C √(5)We know that, √(6 + 2√(5)) 1 = nbsp;√(5+ 1 + 2√(5)) 1 = nbsp;√((√(5))^2+ 1^2 + 2√(5)) 1 = nbsp;√((√(5)+1)^2) 1 = √(5)+1 1 ...

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Byju's Answer

Standard VIII

Mathematics

Distributive Property of Multiplication

The value of ...

Question

The value of $\sqrt{6 + 2 \sqrt{5}} - 1$ is .

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0.0

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\sqrt{5}

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Solution

The correct option is C $\sqrt{5}$
We know that,
$\sqrt{6 + 2 \sqrt{5}} - 1$
$= \sqrt{5 + 1 + 2 \sqrt{5}} - 1$
$= \sqrt{(\sqrt{5})^{2} + 1^{2} + 2 \sqrt{5}} - 1$
$= \sqrt{(\sqrt{5} + 1)^{2}} - 1$
$= \sqrt{5} + 1 - 1$
$= \sqrt{5}$

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Similar questions

Q. The value of

\sqrt{6 + 2 \sqrt{5}} - 1

is .

Q. If n is a multiple of

6

, show that each of the series

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

is equal to zero.

Q. P.T that

(1 \times 2 \times 5) + (2 \times 5 \times 7) + (3 \times 4 \times 9) + . . . . . . i s \frac{n (n + 1) (n + 2) (3 n + 7)}{6}

Q. If

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

then

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

is equal to

Q. Prove that

(1 \times 2 \times 5) + (2 \times 3 \times 7) + (3 \times 4 \times 9) + . . . . .

\frac{n (n + 1) (n + 2) (3 n + 7)}{6}

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The value of √6+2√5−1 is .

The correct option is C √5We know that, √6+2√5−1 =√5+1+2√5−1 =√(√5)2+12+2√5−1 =√(√5+1)2−1 =√5+1−1 =√5

The value of $\sqrt{6 + 2 \sqrt{5}} - 1$ is .

The correct option is C $\sqrt{5}$
We know that,
$\sqrt{6 + 2 \sqrt{5}} - 1$
$= \sqrt{5 + 1 + 2 \sqrt{5}} - 1$
$= \sqrt{(\sqrt{5})^{2} + 1^{2} + 2 \sqrt{5}} - 1$
$= \sqrt{(\sqrt{5} + 1)^{2}} - 1$
$= \sqrt{5} + 1 - 1$
$= \sqrt{5}$