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2 years ago

Which expression best represents the product of two consecutive integers? CLEAR CHECK x?(2+x) x?(x+1) 2?(x+1) 2x+2

Mathematics

1 answer:

Stella [2.4K]2 years ago

4 0

Answer:

<h2>The product of two consecutive integers:</h2><h2>x · (x+1) = x(x + 1)</h2>

Step-by-step explanation:

Consecutive integer:

... -2, -1, 0, 1, 2, 3, ...

Difference between two consecutive integers is equal to 1.

Therefore two consecutive integers is x and x + 1.

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Conjugate/Rational Number?

hram777 [196]

Answer:

1) $\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) $\dfrac{2}{\sqrt{5} }$

$\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

$\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} }$

$-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

$-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} }$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } \times \dfrac{ \sqrt{10} }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50} }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2} }{10} = \dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} }$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6} }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} } = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

6) $\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} }$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}$