Question

choose the following----with solution ....The number (root2 + root3)^10 (root2 - root3)^10 is a a) rational no. b) irrational no. c) prime no. d) negative no.

Answer 1

Let,

$\sqrt{2}+\sqrt{3}=a\\ \\ \sqrt{2}-\sqrt{3}=b$

Thus,

$\begin{aligned}&(\sqrt{2}+\sqrt{3})^{10}(\sqrt{2}-\sqrt{3})^{10}\\ \\ \Longrightarrow\ \ &a^{10}b^{10}\\ \\ \Longrightarrow\ \ &(ab)^{10}\end{aligned}$

What about ab?!

$\begin{aligned}&ab\\ \\ \Longrightarrow\ \ &(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})\\ \\ \Longrightarrow\ \ &(\sqrt{2})^2-(\sqrt{3})^2\\ \\ \Longrightarrow\ \ &2-3\\ \\ \Longrightarrow\ \ &-1\end{aligned}$

Hence,

$(ab)^{10}=(-1)^{10}=1$

Hence the value is 1.

→ 1 is not an irrational number.

→ 1 is not defined as a prime number.

→ 1 is not a negative number.

But 1 is a rational number!

Hence option (a) is the answer.

Answer 2

Answer:

The Answer is 1.

Step-by-step explanation:

Ur explanation is in the attachment.....

We know that when two numbers are multiplied with same powers, then first of all the numbers are multiplied with each other and the power is the same.

So in your question the same process is followed.

First of all both the numbers are multiplied and the power is 10 .

Now the Answer is (-1)^10.

If the power is even then the Answer remains positive.

But if the power is odd, the answer is-ve.

Here the power is even. Hence , -1^10 =1.