Question

Ex 6 si ex 14 va rog frumos este foarte urgent dau coroana

Answer 1

   
[tex]\displaystyle \\ Ex:6) \\ a) \\ \\ \frac{ \sqrt{2} \cdot \sqrt[3]{2} }{ \sqrt{8}}= \frac{ \sqrt{2} \cdot \sqrt[3]{2} }{ \sqrt{2^3}}=\frac{ 2^{ \frac{1}{2} } \cdot 2^\frac{1}{3} }{ 2^\frac{3}{2} }=2^{ \frac{1}{2} +\frac{1}{3} -\frac{3}{2} }=2^ \frac{3+2-9}{6} =2^ \frac{-4}{6}= 2^ {-\frac{2}{3}}[/tex]


[tex]\displaystyle \\ b) \\ \\ \frac{ \sqrt{2 \sqrt{2} } \cdot \sqrt[3]{2} }{ \sqrt[3]{4} } =\frac{ \sqrt{2} \cdot\sqrt{\sqrt{2} } \cdot \sqrt[3]{2} }{ \sqrt[3]{2^4} } = \\ \\ = \frac{2^ \frac{1}{2} \cdot\ 2^ \frac{1}{4} \cdot 2^ \frac{1}{3} }{ 2^ \frac{4}{3} } =2^{\frac{1}{2} +\frac{1}{4} +\frac{1}{3}-\frac{4}{3}}=2^ \frac{6+3+4-12}{12} = 2^ \frac{1}{12} [/tex]


[tex]\displaystyle \\ c) \\ \\ \frac{ \sqrt[3]{ \sqrt{ \sqrt{2} } } }{ \sqrt{2} } = \frac{2^ \frac{1}{2\cdot 2\cdot 3} }{2^ \frac{1}{2} } =\frac{2^ \frac{1}{12} }{2^ \frac{1}{2} } =2^{\frac{1}{12} -\frac{1}{2} }=2^ \frac{1-6}{12}=2^{- \frac{5}{12} } [/tex]




[tex]\displaystyle \\ Ex.14)~~\text{Voi rezolva in ordinea: a, c, b, d} \\ a) ~~\text{Aplicam formula: }(a-b)(a+b)=a^2-b^2 \\ \\ \left(3^ \frac{1}{2}-2^\frac{1}{2} \right)\left(3^\frac{1}{2}+2^ \frac{1}{2} \right)= \left(3^ \frac{1}{2}\right)^2-\left(2^\frac{1}{2} \right)^2 = \\ \\ =3^ {\frac{1}{2}\cdot 2}-2^ {\frac{1}{2}\cdot 2}=3^ {\frac{2}{2}}-2^ {\frac{2}{2}}=3^1-2^1 = 3-2 =1 [/tex]


[tex]\displaystyle \\ c) ~~\text{Aplicam formula: }(a+b)(a-b)=a^2-b^2 \\ \\ \left(x^ \frac{1}{2}+y^\frac{1}{2} \right)\left(x^\frac{1}{2}-y^ \frac{1}{2} \right)= \left(x^ \frac{1}{2}\right)^2-\left(y^\frac{1}{2} \right)^2 = \\ \\ =x^ {\frac{1}{2}\cdot 2}-y^ {\frac{1}{2}\cdot 2}=x^ {\frac{2}{2}}-y^ {\frac{2}{2}}=x^1-y^1 = x-y [/tex]


[tex]\displaystyle \\ b)~~\text{Aplicam formula: }~~(a-b)(a^2+ab+b^2)=a^3-b^3 \\ \\ \left( 3^\frac{1}{3}-2^\frac{1}{3} \right)\left( 3^\frac{2}{3}+6^\frac{1}{3}+2^\frac{2}{3} \right)= \\ \\ =\left( 3^\frac{1}{3}-2^\frac{1}{3} \right)\left(\Big( 3^\frac{1}{3}\Big)^2+3^\frac{1}{3}\cdot 2^\frac{1}{3}+\Big(2^\frac{1}{3}\Big)^2 \right)= \\ \\ \left( 3^\frac{1}{3}\right)^3-\left(2^\frac{1}{3} \right)^3 = 3^{\frac{1}{3}\cdot 3}-2^{\frac{1}{3}\cdot 3}=3^\frac{3}{3} -2^\frac{3}{3} =3^1-2^1 = 3-2=1[/tex]


[tex]\displaystyle \\ d)~~\text{Aplicam formula: }~~(a+b)(a^2-ab+b^2)=a^3+b^3 \\ \\ \left( x^\frac{1}{3}+y^\frac{1}{3} \right)\left( x^\frac{2}{3}-(xy)^\frac{1}{3}+y^\frac{2}{3} \right)= \\ \\ =\left( x^\frac{1}{3}+y^\frac{1}{3} \right)\left(\Big( x^\frac{1}{3}\Big)^2-x^\frac{1}{3}\cdot y^\frac{1}{3}+\Big(y^\frac{1}{3}\Big)^2 \right)= \\ \\ \left( x^\frac{1}{3}\right)^3+\left(y^\frac{1}{3} \right)^3 = x^{\frac{1}{3}\cdot 3}+y^{\frac{1}{3}\cdot 3}=x^\frac{3}{3} +y^\frac{3}{3} =x^1+y^1 = x+y[/tex]