Question

vă rog, dau coroana și 50 de puncte , urgent vă rooooogggg​

Answer 1

1.

$\sqrt{6^2} =6$

$\sqrt{(-5)^2} =\sqrt{-5\cdot(-5)} =\sqrt{25} =5$

$\sqrt{3^8} =\sqrt{3^4\cdot3^4} =3^4=81$

$\sqrt{72} =\sqrt{6^2\cdot2} =6\sqrt{2}$

$\sqrt{81} =\sqrt{9^2} =9$

$\sqrt{3^5} =\sqrt{3^4\cdot3} =3^2\sqrt{3} =9\sqrt{3}$

$\sqrt{7^3\cdot3^2\cdot2^7} =\sqrt{7^2\cdot7\cdot3^2\cdot2^6\cdot2} =7\cdot3\cdot8\cdot\sqrt{7\cdot2} =168\sqrt{14}$

$\sqrt{(-7)^4} =\sqrt{(-7)^2\cdot(-7)^2} =|-7^2|=49$

$\sqrt{9800} =\sqrt{70^2\cdot2} =70\sqrt{2}$

$\sqrt{10125} =\sqrt{45^2\cdot5} =45\sqrt{5}$

$\sqrt{6300} =\sqrt{2^2\cdot5^2\cdot3^2\cdot7} =2\cdot5\cdot3\cdot\sqrt{7} =30\sqrt{7}$

$\sqrt{3150} =\sqrt{2\cdot5^2\cdot3^2\cdot7} =5\cdot3\cdot\sqrt{2\cdot7} =15\sqrt{14}$

2.

$3\sqrt{5} +\sqrt{5} =4\sqrt{5}$

$3\sqrt{5} -\sqrt{5} =2\sqrt{5}$

$3\sqrt{5} \cdot\sqrt{5} =3\sqrt{5} ^2=3\cdot5=15$

$3\sqrt{5} :\sqrt{5} =3$

$7\sqrt{5} -12\sqrt{5} =-5\sqrt{5}$

$3\sqrt{2} \cdot5\sqrt{2} =3\cdot5\cdot\sqrt{2} ^2=15\cdot2=30$

$(10\sqrt{15} ):(5\sqrt{3} )=(10:5)(\sqrt{15} :\sqrt{3} )=2\sqrt{5}$

$(18\sqrt{21} ):(6\sqrt{7} )=(18:6)(\sqrt{21} :\sqrt{7} )=3\sqrt{3}$

3. Un număr negativ la o putere para va fi un număr pozitiv.

$(\sqrt{3} )^4=\sqrt{3^4} =3^2=9$

$(-\sqrt{2} )^2=-\sqrt{2} \cdot(-\sqrt{2} )=\sqrt{2} ^2=2$

$(-\sqrt{2} )^3=-\sqrt{2^3} =-2\sqrt{2}$

$(5\sqrt{7} )^2=5^2\sqrt{7} ^2=25\cdot7=175$

$(-2\sqrt{3} )^3=-2^3\sqrt{3^3} =-8\cdot3\sqrt{3} =-24\sqrt{3}$

$(-\sqrt{11} )^0=1$

$(\sqrt{2} )^{10}=\sqrt{2^{10}} =2^5=32$

$(3\sqrt{2} )^4=3^4\sqrt{2^4} =81\cdot2^2=324$

$(-5\sqrt{3} )^2=5^2\sqrt{3} ^2=25\cdot3=75$

$(11\sqrt{11} )^2=11^2\sqrt{11} ^2=121\cdot11=1331$

$(-5\sqrt{2} )^3=-5^3\sqrt{2} ^3=-125\cdot2\sqrt{2} =-250\sqrt{2}$

$(-\sqrt{7} )^5=-\sqrt{7} ^5=-49\sqrt{7}$

4.

$^{\sqrt{5} )}\frac{3}{\sqrt{5} } =\frac{3\sqrt{5} }{5}$

$^{\sqrt{3} )}\frac{7}{\sqrt{3} } =\frac{7\sqrt{3} }{3}$

$^{\sqrt{2} )}\frac{-5}{3\sqrt{2} } =\frac{-5\sqrt{2} }{3\cdot2} =-\frac{5\sqrt{2} }{6}$

$^{\sqrt{11} )}\frac{11}{-3\sqrt{11} } =\frac{11\sqrt{11} }{-3\cdot11}^{(11} =-\frac{\sqrt{11} }{3}$

$^{\sqrt{5} )}-\frac{15}{2\sqrt{5} } =-\frac{15\sqrt{5} }{2\cdot5} ^{(5}=-\frac{3\sqrt{5} }{2}$

$^{(\sqrt{6} }-\frac{90}{\sqrt{6} } =-\frac{90\sqrt{6} }{6} ^{(6}=-15\sqrt{6}$

5.

a) $5\sqrt{18} +3\sqrt{2} -\sqrt{98} =$

  $5\cdot3\sqrt{2} +3\sqrt{2} -7\sqrt{2} =$

  $15\sqrt{2} +3\sqrt{2} -7\sqrt{2} =$

  $(15+3-7)\sqrt{2} =11\sqrt{2}$

b)$7\sqrt{5} +2\sqrt{125} -\sqrt{20} =$

  $7\sqrt{5} +2\cdot5\sqrt{5} -2\sqrt{5} =$

  $7\sqrt{5} +10\sqrt{5} -2\sqrt{5} =$

  $(7+10-2)\sqrt{5} =15\sqrt{5}$

c)$4\sqrt{7} -3\sqrt{28} +\sqrt{7} =$

  $4\sqrt{7} -3\cdot2\sqrt{7} -\sqrt{7} =$

  $4\sqrt{7} -6\sqrt{7} -\sqrt{7} =$

  $(4-6-1)\sqrt{7} =-3\sqrt{7}$

6.

$^{\sqrt{3})} \frac{6}{\sqrt{3} } +2\sqrt{3} -\sqrt{27} =$

$\frac{6\sqrt{3} }{3} +2\sqrt{3} -3\sqrt{3} =$

$2\sqrt{3} +2\sqrt{3} -3\sqrt{3} =$

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

__________________

$^{\sqrt{2})} \frac{10}{\sqrt{2} } +^{\sqrt{6})} \frac{12\sqrt{2} }{\sqrt{6} } -2\sqrt{50} =$

$\frac{10\sqrt{2} }{2} +\frac{12\sqrt{12} }{6} -2\cdot5\sqrt{2} =$

$5\sqrt{2} +2\cdot2\sqrt{3} -10\sqrt{2} =$

$5\sqrt{2} +4\sqrt{3} -10\sqrt{2} =$

$-5\sqrt{2} +4\sqrt{3}$

____________________

$^{\sqrt{2})} \frac{5 }{\sqrt{2} } +^{\sqrt{2})} \frac{7}{3\sqrt{2} } -^{\sqrt{6})} \frac{2\sqrt{3} }{\sqrt{6} } =$

$\frac{5\sqrt{2} }{2} +\frac{7\sqrt{2} }{3\cdot2} -\frac{2\sqrt{18} }{6} =$

$^{3)}\frac{5\sqrt{2} }{2} +\frac{7\sqrt{2} }{6} -\frac{6\sqrt{2} }{6} =$

$\frac{15\sqrt{2}+7\sqrt{2} -6\sqrt{2} }{6} =\frac{16\sqrt{2} }{6} ^{(2}=\frac{8\sqrt{2} }{3}$

7.

$m_g=\sqrt{a\cdot b}$

$m_g=\sqrt{6\cdot8} =\sqrt{48} =4\sqrt{3}$

$m_g=\sqrt{10\cdot15} =\sqrt{150} =5\sqrt{6}$

$m_g=\sqrt{\sqrt{3} \cdot\sqrt{27} } =\sqrt{\sqrt{3}\cdot3\sqrt{3} } =\sqrt{3\cdot3} =3$

$m_g=\sqrt{2\sqrt{45} \cdot3\sqrt{20} } =\sqrt{2\cdot3\sqrt{5} \cdot3\cdot2\sqrt{5} }$

$=\sqrt{2^2\cdot3^2\cdot5} =6\sqrt{5}$

$m_g=\sqrt{7\sqrt{5}\cdot\sqrt{125} } =\sqrt{7\sqrt{5}\cdot5\sqrt{5} } =\sqrt{7\cdot5^2} =5\sqrt{7}$

$\succ \supset---\Delta Triunghiul_1 \Delta---\subset \prec$