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

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

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

Answer 2

спасибо бальшое