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Pe: 2020-09-21T21:25:41+03:00 In: In:

Va rog frumos ajutati.m a

Va rog frumos ajutati.ma

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  1. Suport

    1.1)\sqrt8-\sqrt{32}+\sqrt{128}=\sqrt{2^3}-\sqrt{2^5}+\sqrt{2^7}=2\sqrt2-2^2\sqrt2+2^3\sqrt2=2\sqrt2-4\sqrt2+8\sqrt2=(2-4+8)\sqrt2=6\sqrt2
    2)\sqrt5+\sqrt{125}-\sqrt{50}=\sqrt5+\sqrt{5^3}-\sqrt{5\cdot5^2}=\sqrt5+5\sqrt5-5\sqrt2=6\sqrt5-5\sqrt2
    3)\sqrt{243}-\sqrt{375}+\sqrt{48}=\sqrt{3^5}-\sqrt{3\cdot5^3}+\sqrt{3\cdot4^2}=3^2\sqrt3-5\sqrt{3\cdot5}+4\sqrt3=9\sqrt3-5\sqrt15+4\sqrt3=13\sqrt3-5\sqrt15

    2.A compara 2 numere pozitive este acelasi lucru cu a le compara patratele.
    1)(5\sqrt3)^2=25\cdot3=75
    (3\sqrt6)^2=9\cdot6=54
    75>54, deci primul numar este mai mare.
    2)(4\sqrt2)^2=16\cdot2=32
    (9\sqrt4)^2=81\cdot4=324
    324>32, deci al 2-lea numar este mai mare.
    3)(7\sqrt5)^2=49\cdot5=245
    (5\sqrt7)^2=25\cdot7=175
    245>175, deci primul numar este cel mai mare.

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