Question

Решите пж срочно надо, умоляю:

Answer 1

№ 7
1.
  $x leq 0; y geq 0 \ \ sqrt{25y} = 5 sqrt{y} \sqrt{0,01x^2} =0,1|x|=-0,1x \ sqrt{2,89x^6y^5} = 1,7|x^3|y^2 sqrt{y} =-1,7x^3y^2 sqrt{y}$

2.
$5 sqrt{2} = sqrt{50} \ 7x sqrt{xy} = sqrt{49x^3y} \ frac{6a sqrt{x} }{ sqrt{3a} } = sqrt{ frac{36a^2x}{3a} } = sqrt{12ax}$

3.
$sqrt{8} - sqrt{18} + sqrt{3} =2 sqrt{2} -3 sqrt{2} + sqrt{3} = - sqrt{2} + sqrt{3} \ \ sqrt{20} -2 sqrt{2}( sqrt{10} -4 sqrt{2} )=2 sqrt{5} -4sqrt{5} +16=-2 sqrt{5} +16$

4.
$frac{7}{2 sqrt{7} } = frac{ sqrt{7} }{2} \ \ frac{2x+ sqrt{x} }{4 sqrt{x} } = frac{ sqrt{x} (2 sqrt{x} +1)}{4 sqrt{x} } = frac{2 sqrt{x} +1}{4} \ \ frac{25 - x}{5+ sqrt{x} } = frac{(5- sqrt{x} )(5+ sqrt{x} )}{5+ sqrt{x} } =5- sqrt{x} \ \ frac{27-a sqrt{a} }{3- sqrt{a} } = frac{(3- sqrt{a} )(a+3 sqrt{a} +9)}{3- sqrt{a} } = a+3 sqrt{a}+9$

5.
$frac{2}{ sqrt{6} } = frac{2 sqrt{6} }{6} = frac{sqrt{6} }{3} \ \ frac{1}{3 sqrt{7} } = frac{ sqrt{7} }{21}$

$frac{4}{ sqrt{7}- sqrt{3} } = frac{4( sqrt{7}+ sqrt{3} )}{( sqrt{7} - sqrt{3} )( sqrt{7} + sqrt{3} )} = frac{4( sqrt{7} + sqrt{3}) }{7-3} = sqrt{7} + sqrt{3} \ \ frac{11}{5- sqrt{14} } - frac{2}{4- sqrt{14} } = frac{11(5+ sqrt{14} )}{25-14} - frac{2(4+ sqrt{14} )}{16-14} =5+ sqrt{14} -4 - sqrt{14} =1$

6.
$frac{ sqrt{a^3}- sqrt{b^3} }{a-b} - frac{ sqrt{a^3}+ sqrt{b^3} }{a+b}= frac{( sqrt{a^3}- sqrt{b^3})(a+b)-( sqrt{a^3}+ sqrt{b^3})(a-b)}{(a-b)(a+b)} = \ \ = frac{a^2 sqrt{a}-ab sqrt{b} +ab sqrt{a} -b^2 sqrt{b} -a^2 sqrt{a} -ab sqrt{b} +ab sqrt{a} +b^2 sqrt{b} }{(a-b)(a+b)} = \ \ = frac{2ab sqrt{a} -2ab sqrt{b} }{(a-b)(a+b)} = frac{2ab( sqrt{a}- sqrt{b} )}{(a-b)(a+b)}$

$a= sqrt{3} - sqrt{2} ;b= sqrt{3} + sqrt{2} \ \ frac{2ab( sqrt{a}- sqrt{b} )}{(a-b)(a+b)}= frac{2( sqrt{3} - sqrt{2})(sqrt{3} + sqrt{2})( sqrt{sqrt{3} - sqrt{2}}- sqrt{sqrt{3} + sqrt{2}} )}{(sqrt{3} - sqrt{2}- sqrt{3}- sqrt{2} )(sqrt{3} - sqrt{2}+ sqrt{3} + sqrt{2} )}= \ \ = frac{2*1*( sqrt{sqrt{3} - sqrt{2}}-sqrt{sqrt{3} + sqrt{2}})}{-2 sqrt{2}*2 sqrt{3} } =- frac{ sqrt{sqrt{3} - sqrt{2}}-sqrt{sqrt{3} + sqrt{2}}}{2 sqrt{2} sqrt{3} } =$
$=- frac{ sqrt{sqrt{3} - sqrt{2}}}{2 sqrt{2} sqrt{3} }+ frac{ sqrt{sqrt{3} +sqrt{2}}}{2 sqrt{2} sqrt{3} } =- frac{1}{2} sqrt{ frac{ sqrt{3} - sqrt{2} }{6 } } + frac{1}{2} sqrt{ frac{ sqrt{3} + sqrt{2} }{6} }$

7.
$sqrt{10} \ 2 sqrt{3} = sqrt{12} \ 3= sqrt{9} 3; sqrt{10} ;2 sqrt{3}$

8.
$( frac{x sqrt{x} -y sqrt{y} }{ sqrt{x} - sqrt{y} } + sqrt{xy} ):( frac{x-y}{ sqrt{x} - sqrt{y} } )^2= \ \ =(frac{(x sqrt{x} -y sqrt{y} )( sqrt{x} + sqrt{y} )}{( sqrt{x} - sqrt{y})( sqrt{x} + sqrt{y} ) } + sqrt{xy} ):( frac{ (sqrt{x} - sqrt{y}) ( sqrt{x} + sqrt{y} )}{ sqrt{x} - sqrt{y} } )^2=$
$=(frac{x^2+x sqrt{xy}-y sqrt{xy}-y^2}{x-y} + sqrt{xy} ):( sqrt{x} + sqrt{y} )^2= \ \ =(frac{(x-y)(x+y)+sqrt{xy}(x-y)}{x-y} + sqrt{xy} ):( sqrt{x} + sqrt{y} )^2= \ \ =(x + y + sqrt{xy}+ sqrt{xy} ):( sqrt{x} + sqrt{y} )^2= \ \ =( sqrt{x} + sqrt{y} )^2:( sqrt{x}+ sqrt{y} )^2=1$