Proszę o rozwiązania wraz z obliczeniami.1. Korzystając w odpowiedniego wzoru skróconego mnożenia, u.... Question from @nezumi1

September 2018 1 29 Report

Proszę o rozwiązania wraz z obliczeniami.

1. Korzystając w odpowiedniego wzoru skróconego mnożenia, usuń niewymierność z mianownika liczby:

a) $\frac{3}{ \sqrt[3]{4} + 2 }$

b) $\frac{2}{ \sqrt[3]{25}+ \sqrt[3]{15}+ \sqrt[3]{9} }$

2. Oblicz sumę:

a) $\frac{1}{ \sqrt{1}+ \sqrt{2} }+ \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+ \sqrt{4} }$

b) $\frac{1}{ \sqrt{1}+ \sqrt{2} }+ \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+ \sqrt{4} } ... \frac{1}{ \sqrt{99}+ \sqrt{100} }$

Roma

Zad. 1

Skorzystamy ze wzoru: (a + b)(a² - ab + b²) = a³ + b³

$\frac{3}{\sqrt[3]{4}+2} = \frac{3}{\sqrt[3]{4} + 2} \cdot \frac{(\sqrt[3]{4})^2 - \sqrt[3]{4} \cdot 2+ 2^2}{(\sqrt[3]{4})^2 - \sqrt[3]{4} \cdot 2+ 2^2} =$

$= \frac{3 \cdot ( \sqrt[3]{16}-2 \cdot \sqrt[3]{4}+4) }{ (\sqrt[3]{4})^3+2^3} =\frac{3 \cdot (\sqrt[3]{8 \cdot 2}-2 \cdot \sqrt[3]{4}+4)}{4+8} =\frac{3 \cdot (2\sqrt[3]{2}-2 \cdot \sqrt[3]{4}+4)}{12}=$

$=\frac{2\sqrt[3]{2}-2 \cdot \sqrt[3]{4}+4}{4}=\frac{2 \cdot (\sqrt[3]{2}-\sqrt[3]{4}+2)}{4}= \frac{\sqrt[3]{2}-\sqrt[3]{4}+2}{2}$

Skorzystamy ze wzoru: (a - b)(a² + ab + b²) = a³ - b³

$\frac{2}{ \sqrt[3]{25}+ \sqrt[3]{15}+ \sqrt[3]{9} } = \frac{2}{(\sqrt[3]{5})^2 + \sqrt[3]{5} \cdot \sqrt[3]{3} +( \sqrt[3]{3})^2 } =$

$=\frac{2}{(\sqrt[3]{5})^2 + \sqrt[3]{5} \cdot \sqrt[3]{3} +( \sqrt[3]{3})^2} \cdot \frac{\sqrt[3]{5}-\sqrt[3]{3}}{\sqrt[3]{5}-\sqrt[3]{3}} =$

$=\frac{2 \cdot (\sqrt[3]{5}-\sqrt[3]{3})}{(\sqrt[3]{5})^3-(\sqrt[3]{3})^3}} =\frac{2 \cdot (\sqrt[3]{5}-\sqrt[3]{3})}{5-3} =\frac{2 \cdot (\sqrt[3]{5}-\sqrt[3]{3})}{2} = \sqrt[3]{5}-\sqrt[3]{3}$

Zad. 2

$\frac{1}{ \sqrt{1}+ \sqrt{2} }+ \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+ \sqrt{4} } = \frac{\sqrt{1}- \sqrt{2}}{ (\sqrt{1}+ \sqrt{2})(\sqrt{1}- \sqrt{2})} +$

$+ \frac{\sqrt{2}- \sqrt{3}}{ (\sqrt{2}+ \sqrt{3})(\sqrt{2}- \sqrt{3})} + \frac{\sqrt{3}- \sqrt{4}}{ (\sqrt{3}+ \sqrt{4})(\sqrt{3}- \sqrt{4})} = \frac{\sqrt{1}- \sqrt{2}}{1-2} +$

$+ \frac{\sqrt{2}- \sqrt{3}}{2-3} + \frac{\sqrt{3}- \sqrt{4}}{3-4} = \frac{1- \sqrt{2}}{-1} + \frac{\sqrt{2}- \sqrt{3}}{-1} + \frac{\sqrt{3}- \sqrt{4}}{-1} =$

$= \frac{1- \sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3} -\sqrt{4}}{-1} = \frac{1-\sqrt{4}}{-1} = \frac{1-2}{-1} = \frac{-1}{-1} = 1$

$\frac{1}{ \sqrt{1}+ \sqrt{2} }+ \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+ \sqrt{4} } +...+ \frac{1}{ (\sqrt{99}+ \sqrt{100}} =$

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

$+ \frac{\sqrt{3}- \sqrt{4}}{ (\sqrt{3}+ \sqrt{4})(\sqrt{3}- \sqrt{4})} + ... + \frac{\sqrt{99}- \sqrt{100}}{ (\sqrt{99}+ \sqrt{100})(\sqrt{99}- \sqrt{100})} =$

$= \frac{\sqrt{1}- \sqrt{2}}{1-2} + \frac{\sqrt{2}- \sqrt{3}}{2-3} + \frac{\sqrt{3}- \sqrt{4}}{3-4} + ... + \frac{\sqrt{99}- \sqrt{100}}{99-100} = \frac{1- \sqrt{2}}{-1} +$