Question

X=radical din 19 supra 2,(1) + radical din 20 supra 2,(2) +radical din 21 supra 2, (3)+....... Radical din 26 supra 2,(8)​

Answer 1

 

N-ai folosit paranteze pentru a se intelege daca numitorul este sau nu sub radical, dar tinand cont de datele problemei consider ca DA, si numitorul este sub radical.

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$\displaystyle\bf\\Rezolvare:\\\\x=\sqrt{\frac{19}{2,\!(1)}}+\sqrt{\frac{20}{2,\!(2)}}+\sqrt{\frac{21}{2,\!(3)}}+\sqrt{\frac{22}{2,\!(4)}}+\\\\+\sqrt{\frac{23}{2,\!(5)}}+\sqrt{\frac{24}{2,\!(6)}}+\sqrt{\frac{25}{2,\!(7)}}+\sqrt{\frac{26}{2,\!(8)}}$

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$\displaystyle\bf\\x=\sqrt{\frac{19}{\dfrac{21-2}{9}}}+\sqrt{\frac{20}{\dfrac{22-2}{9}}}+\sqrt{\frac{21}{\dfrac{23-2}{9}}}+\sqrt{\frac{22}{\dfrac{24-2}{9}}}+\\\\+\sqrt{\frac{23}{\dfrac{25-2}{9}}}+\sqrt{\frac{24}{\dfrac{26-2}{9}}}+\sqrt{\frac{25}{\dfrac{27-2}{9}}}+\sqrt{\frac{26}{\dfrac{28-2}{9}}}$

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$\displaystyle\bf\\x=\sqrt{\frac{19}{\dfrac{19}{9}}}+\sqrt{\frac{20}{\dfrac{20}{9}}}+\sqrt{\frac{21}{\dfrac{21}{9}}}+\sqrt{\frac{22}{\dfrac{22}{9}}}+\\\\+\sqrt{\frac{23}{\dfrac{23}{9}}}+\sqrt{\frac{24}{\dfrac{24}{9}}}+\sqrt{\frac{25}{\dfrac{25}{9}}}+\sqrt{\frac{26}{\dfrac{26}{9}}}$

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$\displaystyle\bf\\x=\sqrt{\frac{19\times9}{19}}+\sqrt{\frac{20\times9}{20}}+\sqrt{\frac{21\times9}{21}}}+\sqrt{\frac{22\times9}{22}}+\\\\+\sqrt{\frac{23\times9}{23}}+\sqrt{\frac{24\times9}{24}}+\sqrt{\frac{25\times9}{25}}+\sqrt{\frac{26\times9}{26}}\\\\\\x=\sqrt{9}+\sqrt{9}+\sqrt{9}+\sqrt{9}+\sqrt{9}+\sqrt{9}+\sqrt{9}+\sqrt{9}\\\\x=3+3+3+3+3+3+3+3\\\\x=8\times3\\\\\boxed{\bf x=24}$