Question

Алгебра заранее 667(а,в) 669(а, б) ​

Answer 1

667

a) 0,1

в) 218,7

669

а) 12 104

б) -4

Объяснение:

667

a)

$\frac{\sqrt{18}-\sqrt{8} }{2\sqrt{50} }=\frac{\sqrt{9 * 2}-\sqrt{4 * 2} }{2\sqrt{25*2} }=\frac{\sqrt{9}*\sqrt{2} -\sqrt{4}*\sqrt{2} }{2\sqrt{25}*\sqrt{2} }=\frac{3\sqrt{2} -2\sqrt{2} }{2*5\sqrt{2} }=\frac{\sqrt{2}(3 -2) }{10\sqrt{2} }=\frac{1}{10} =0,1$

в)

$(2\sqrt{55} )^{2} -\sqrt{3} *\sqrt{1,08} +\frac{\sqrt{7}}{\sqrt{28}} =2^{2} *(\sqrt{55})^{2} -\sqrt{3}*\sqrt{3*0,36} + \frac{\sqrt{7}}{\sqrt{4*7}}=4*55-\sqrt{3}*\sqrt{3}*\sqrt{0,36} +\frac{\sqrt{7}}{\sqrt{4}*\sqrt{7}}=220-(\sqrt{3})^{2} *0,6+\frac{1}{2} =220-3*0,6+0,5=220,5-1,8=218,7$

669

а)

$(1+\sqrt{2017} )^{3} +(1-\sqrt{2017} )^{3}=(1^{3} +3*1^{2} *\sqrt{2017} +3*1*(\sqrt{2017} )^{2} +(\sqrt{2017} )^{3} )+(1^{3} -3*1^{2} *\sqrt{2017} +3*1*(\sqrt{2017} )^{2} -(\sqrt{2017} )^{3})=1 +3*2017 +3*2017 +(\sqrt{2017} )^{3} +1 -3*2017} +3*2017 -(\sqrt{2017} )^{3}=2+3*2017 +3*2017=2+2(3*2017)= 2+2*6051=2+12102=12104$

б)

$\sqrt{10-4\sqrt{6} } -\sqrt{10+4\sqrt{6} }$

$4\sqrt{6}=2*\sqrt{6} *2$

$10=6+4=(\sqrt{6} )^{2} +2^{2}$

$\sqrt{10+4\sqrt{6} } = \sqrt{(\sqrt{6} )^{2}-2*\sqrt{6}*2+2^{2}} =\sqrt{(\sqrt{6}-2 )^{2} } =\sqrt{6} -2$

$\sqrt{10+4\sqrt{6} } = \sqrt{(\sqrt{6} )^{2}+2*\sqrt{6}*2+2^{2}} =\sqrt{(\sqrt{6}+2 )^{2} } =\sqrt{6} +2$

$\sqrt{10-4\sqrt{6} } -\sqrt{10+4\sqrt{6} }=(\sqrt{6} -2)-(\sqrt{6} +2)=\sqrt{6} -2-\sqrt{6}-2=-2-2=-4$

Формулы:

$\sqrt{a*b} =\sqrt{a} *\sqrt{b}$

$(a*b)^{2} =a^{2} *b^{2}$

$(\sqrt{a}) ^{2} =\sqrt{a} ^{2} =a$

$(a+b)^{3} =a^{3} +3*a^{2} *b+3*a*b^{2} +b^{3}$

$(a-b)^{3} =a^{3} -3*a^{2} *b+3*a*b^{2} -b^{3}$

$(a-b)^{2} =a^{2} -2*a*b+b^{2}$

$(a+b)^{2} =a^{2} +2*a*b+b^{2}$

$\sqrt{a^{2}} =a$