Question

Nu se inlocuieste radical din 2 cu 2^1/2?Nu inteleg cum rezultatul exercitiului este 3/2

Answer 1

[tex]\it \dfrac{4}{\sqrt2}=\dfrac{2\cdot2}{\sqrt2} =\dfrac{2\cdot\sqrt2\cdot\sqrt2}{\sqrt2}=2\sqrt2 =2^1\cdot2^{\frac{1}{2}} =2^{1+\frac{1}{2}} =2^{\frac{3}{2}} \\\;\\ \\\;\\ Expresia\ \ devine:\ log_2 2^{\frac{3}{2}} =\dfrac{3}{2}log_2 2=\dfrac{3}{2}\cdot1=\dfrac{3}{2} =1,5[/tex]


Observație:

[tex]\it \sqrt2\cdot\sqrt2=(\sqrt{2})^2=2 \\\;\\ Sau:\\\;\\ \sqrt2\cdot\sqrt2=\sqrt{2\cdot2} =\sqrt4=2 [/tex]

Prin urmare, 2 se poate scrie 2= √2·√2

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R₂:

[tex]\it \log_2\dfrac{4}{\sqrt2} =log_2 4-log_2{\sqrt2} =log_2 2^2-log_2 2^{\frac{1}{2}} =2log_2 2-\dfrac{1}{2}log_2 2 = \\\;\\ \\\;\\ =2-\dfrac{1}{2} =\dfrac{^{2)}2}{\ 1}-\dfrac{1}{2} =\dfrac{4-1}{2} =\dfrac{3}{2}[/tex]

Observație:

$\it \dfrac{4}{\sqrt2} =\dfrac{2^2}{2^{\frac{1}{2}}} =2^{2-\frac{1}{2}} =2^{\frac{3}{2}}$

Answer 2

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