Question

памогите найти квадратные корни

Answer 1

1. $\frac{36}{(6*\sqrt{2})^{2}  } =\frac{36}{6^{2}*(\sqrt{2})^{2}   } =\frac{36}{36*2} =\frac{1}{2}$

2$\sqrt{\frac{9}{16} } +\sqrt{\frac{25}{36} } =\frac{\sqrt{9} }{\sqrt{16}} +\frac{\sqrt{25}}{\sqrt{36} } =\frac{3}{4} +\frac{5}{6} =\frac{9}{12} +\frac{10}{12} =\frac{19}{12}} =1\frac{7}{12}$

3 $(\sqrt{2} +\sqrt{3} )(\sqrt{2} -\sqrt{3} )=(\sqrt{2} )^{2} -(\sqrt{3} )^{2} =2-3=-1$

4 $(\sqrt{12} +\sqrt{3} )^{2} =(\sqrt{12} )^{2} +2*\sqrt{12} *\sqrt{3} +(\sqrt{3} )^{2} =12+2*\sqrt{4} *\sqrt{3} *\sqrt{3} +3=12+12+3=27$

5 $0.5*\sqrt{0.04} +\frac{1}{6} *\sqrt{144} =0.5*0.2+\frac{1}{6} *12=0.1+2=2.1$

6 $10*\sqrt{3} -4\sqrt{48} -\sqrt{27} =10*\sqrt{3} -4*\sqrt{16*3} -\sqrt{9*3} =10*\sqrt{3} -4*\sqrt{16} *\sqrt{3} -\sqrt{9} *\sqrt{3} =10\sqrt{3} -16\sqrt{3} -3\sqrt{3} =-9\sqrt{3}$

7 $\sqrt{8} *\sqrt{18} =\sqrt{4*2} *\sqrt{9*2} =\sqrt{4} *\sqrt{2} *\sqrt{9} *\sqrt{2} =2*\sqrt{2} *3*\sqrt{2} =6*(\sqrt{2)} )^{2} =6*2=12$

8 $\sqrt{2^{4} *5^{6} } =\sqrt{2^{4} }*\sqrt{5^{6} } =2^{2}  *5^{3} =4*125=500$