Wzory skróconego mnożenia Dla dowolnych liczb $a$ i $b$ zachodzi: $\left(a-b\right)\left(a+b\right)=a^2-b^2$ ${\left(a+b\right)}^2=a^2+2ab+b^2$ ${\left(a-b\right)}^2=a^2-2ab+b^2$

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Chcemy uprościć podane wyrażenia. W tym celu skorzystamy ze wzorów skróconego mnożenia.

a)

$\left(2-3\sqrt{3}\right)\left(2-3\sqrt{3}\right)={\left(2-3\sqrt{3}\right)}^2=2^2-2\cdot 2\cdot 3\sqrt{3}+{\left(3\sqrt{3}\right)}^2=4-12\sqrt{3}+9\cdot 3=31-12\sqrt{3}$

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b)

$\left(7+5\sqrt{2}\right)\left(5\sqrt{2}+7\right)={\left(7+5\sqrt{2}\right)}^2=7^2+2\cdot 5\cdot 7\sqrt{2}+{\left(5\sqrt{2}\right)}^2=49+70\sqrt{2}+25\cdot 2=99+70\sqrt{2}$

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c)

$\left(4\sqrt{5}+\sqrt{3}\right)\left(4\sqrt{5}-\sqrt{3}\right)={\left(4\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2=16\cdot 5-3=77$

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d)

$\left(\sqrt{8}+2\sqrt{6}\right)\left(2\sqrt{6}-\sqrt{8}\right)={\left(2\sqrt{6}\right)}^2-{\left(\sqrt{8}\right)}^2=4\cdot 6-8=16$

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e)

${\left(\sqrt{6}-3\sqrt{3}\right)}^2={\left(\sqrt{6}\right)}^2-2\cdot 3\sqrt{3}\cdot \sqrt{6}+{\left(3\sqrt{3}\right)}^2=6-6\sqrt{18}+9\cdot 3=6+27-6\sqrt{9\cdot 2}=33-6\cdot 3\sqrt{2}=33-18\sqrt{2}$

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f)

${\left(2\sqrt{32}-3\sqrt{2}\right)}^2={\left(2\sqrt{16\cdot 2}-3\sqrt{2}\right)}^2={\left(2\cdot 4\sqrt{2}-3\sqrt{2}\right)}^2={\left(8\sqrt{2}-3\sqrt{2}\right)}^2={\left(5\sqrt{2}\right)}^2=5^2\cdot 2=50$