Find the product-3-sqrt-2-2-sqrt-3-3-sqrt-2-2-sqrt-3-

Question

Toppr Admin · Accepted Answer

We have-3-x221A-2-2-x221A-3-3-x2212-x221A-2-2-x2212-x221A-3-3-x221A-2-3-x2212-x221A-2-2-x221A-3-2-x2212-x221A-3-We know that-a2-x2212-b2-a-b-a-x2212-b-Therefore-9-x2212-2-4-x2212-3-7-xD7-1-7Hence- the product is 7-

Find the product.
$(3 + \sqrt{2}) (2 + \sqrt{3}) (3 - \sqrt{2}) (2 - \sqrt{3})$

We have,
$= (3 + \sqrt{2}) (2 + \sqrt{3}) (3 - \sqrt{2}) (2 - \sqrt{3})$
$= (3 + \sqrt{2}) (3 - \sqrt{2}) (2 + \sqrt{3}) (2 - \sqrt{3})$

We know that,
$a^{2} - b^{2} = (a + b) (a - b)$

Therefore,
$= (9 - 2) (4 - 3)$
$= 7 \times 1$
$= 7$

Hence, the product is $7.$

Find the product.(3+√2)(2+√3)(3−√2)(2−√3)

We have,=(3+√2)(2+√3)(3−√2)(2−√3)=(3+√2)(3−√2)(2+√3)(2−√3)We know that,a2−b2=(a+b)(a−b)Therefore,=(9−2)(4−3)=7×1=7Hence, the product is 7.

Find the product.
$(3 + \sqrt{2}) (2 + \sqrt{3}) (3 - \sqrt{2}) (2 - \sqrt{3})$

We have,
$= (3 + \sqrt{2}) (2 + \sqrt{3}) (3 - \sqrt{2}) (2 - \sqrt{3})$
$= (3 + \sqrt{2}) (3 - \sqrt{2}) (2 + \sqrt{3}) (2 - \sqrt{3})$

We know that,
$a^{2} - b^{2} = (a + b) (a - b)$

Therefore,
$= (9 - 2) (4 - 3)$
$= 7 \times 1$
$= 7$

Hence, the product is $7.$