памагите задания 4,11. Created by alekseenkoanastasia2. algebra-ru

Ответ:1)2)3)4)5)6)Объяснение:1)2)3)4) 5)6)

памагите задания 4,11...

alekseenkoanastasia2 @alekseenkoanastasia2

July 2022 1 8 Report

памагите задания 4,11

Answers & Comments

Banana32411

Ответ:

$\frac{1}{a + \sqrt{3} }$

$\frac{1}{b - \sqrt{7} }$

$\frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{3} }$

$\frac{a}{a + \sqrt{5} }$

$\frac{1}{3 \sqrt{x} + 2 \sqrt{a} }$

$\frac{1}{7 - \sqrt{3c} }$

Объяснение:

$\frac{a - \sqrt{3} }{a {}^{2} - 3 } = \frac{a - \sqrt{3} }{(a - \sqrt{3) \times (a + \sqrt{3)} } } = \frac{1}{a + \sqrt{3} }$

$\frac{b + \sqrt{7} }{b {}^{2} - 7 } = \frac{b + \sqrt{7} }{(b - \sqrt{7}) \times (b + \sqrt{7)} } = \frac{1}{b - \sqrt{7} }$

$\frac{a + \sqrt{3a} }{a - 3} = \frac{ \sqrt{a}( \sqrt{a} + \sqrt{a}) }{( \sqrt{a} - \sqrt{3}) \times ( \sqrt{ a } + \sqrt{3} ) } = \frac{ \sqrt{a} }{ \sqrt{a} - \sqrt{3} }$

$\frac{a {}^{2} - a \sqrt{5} }{a {}^{2} - 5 } = \frac{a {}^{2} - a \sqrt{5} }{(a - \sqrt{5}) \times (a + \sqrt{5}) } = \frac{a \times ( - \sqrt{5}) }{(a - \sqrt{5}) \times (a + \sqrt{5} )} = \frac{a}{a + \sqrt{5} }$

$\frac{3 \sqrt{x} - 2 \sqrt{a} }{9x - 4a} = \frac{3 \sqrt{x} - 2 \sqrt{a} }{(3 \sqrt{x} - 2 \sqrt{a}) \times (3 \sqrt{x} + 2 \sqrt{a)} } = \frac{1}{3 \sqrt{x} + 2 \sqrt{a} }$

$\frac{7 + \sqrt{3c} }{49 - 3c} = \frac{7 + \sqrt{3c} }{(7 - \sqrt{3c) \times (7 + \sqrt{3c)} } } = \frac{1}{7 - \sqrt{3c} }$

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