Antekstryczynski581
Rozwiązane

Oblicz.
[tex]3\sqrt{5} +4\sqrt{5}[/tex]
[tex]\sqrt{33} *\sqrt3{\frac{2}{3} }[/tex]
[tex]\sqrt{300} -7\sqrt{3}[/tex]
[tex]3\sqrt{6} *\frac{1}{2} \sqrt{3}[/tex]
[tex](\frac{4\sqrt{6} }{3} )^{2}[/tex]
na jutro pls



Odpowiedź :

Tojamiki

Odpowiedź:

3√5 + 4√5 = 7√5

√33 * √(3 2/3) = √33 * √11/3 = √(33*11/3) = √121 = 11

√300 - 7√3 = √(3*100) - 7√3 = 10√3 - 7√3 = 3√3

3√6 * 1/2√3 = 3/2√(6*3) = 3/2√18 = 3/2√(9*2) = 3/2 * 3√2 = 9/2√2 =

= 4 1/2√2

(4√6/3)² = 16/9 * 6 = 16/3 * 2 = 32/3 = 10 2/3

Szczegółowe wyjaśnienie:

Djpancernikfotk

[tex]3\sqrt{5}+4\sqrt{5}=7\sqrt{5}[/tex]

[tex]\sqrt{33}\cdot\sqrt{3\dfrac{2}{3}}=\sqrt{33\cdot\dfrac{11}{3}}=\sqrt{11\cdot\dfrac{11}{1}}=\sqrt{11^2}=11[/tex]

[tex]\sqrt{300}-7\sqrt{3}=\sqrt{100\cdot3}-7\sqrt{3}=10\sqrt{3}-7\sqrt{3}=3\sqrt{3}[/tex]

[tex]3\sqrt{6}\cdot\dfrac{1}{2}\sqrt{3}=\dfrac{3}{2}\sqrt{18}=\dfrac{3}{2}\sqrt{9\cdot2}=\dfrac{3}{2}\cdot3\sqrt{2}=\dfrac{9}{2}\sqrt{2}[/tex]

[tex]\left(\dfrac{4\sqrt{6}}{3}\right)^2=\dfrac{16\cdot6}{9}=\dfrac{16\cdot2}{3}=\dfrac{32}{3}=10\dfrac{2}{3}[/tex]

[tex]3\sqrt{6}\cdot\dfrac{1}{2}\sqrt{2}=\dfrac{3}{2}\sqrt{12}=\dfrac{3}{2}\sqrt{4\cdot3}=\dfrac{3}{2}\cdot2\sqrt{3}=\dfrac{3}{1}\cdot1\sqrt{3}=3\sqrt{3}[/tex]

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