Question

ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО! СДЕЛАЙТЕ ПОЛНОСТЬЮ ВСЕ ЗАДАНИЕ!

Answer 1

$\displaystyle\bf\\1)\\\\\sqrt[3]{7-\sqrt{22} } \cdot\sqrt[3]{7+\sqrt{22} }=\sqrt[3]{(7-\sqrt{22})(7+\sqrt{22} } =\sqrt[3]{7^{2} -(\sqrt{22} )^{2} } =\\\\\\=\sqrt[3]{49-22} =\sqrt[3]{27} =\sqrt[3]{3^{3} } =3\\\\\\2)\\\\\frac{1}{4+2\sqrt{3} } +\frac{1}{4-2\sqrt{3} }=\frac{4-2\sqrt{3}+4+2\sqrt{3} }{(4+2\sqrt{3})(4-2\sqrt{3} )} =\frac{8}{4^{2} -(2\sqrt{3} )^{2} } =\\\\\\=\frac{8}{16-12} =\frac{8}{4} =2$

$\displaystyle\bf\\3)\\\\\frac{3}{6-2\sqrt{6} } +\frac{3}{6+2\sqrt{6} } =\frac{3\cdot(6+2\sqrt{6} )+3\cdot(6-2\sqrt{6}) }{(6-2\sqrt{6})(6+2\sqrt{6}) } =\\\\\\=\frac{18+6\sqrt{6} +18-6\sqrt{6} }{6^{2} -(2\sqrt{6})^{2} } =\frac{36}{36-24} =\frac{36}{10} =3,6\\\\\\4)\\\\\sqrt{3} +2+\frac{1}{2+\sqrt{3} } =\frac{(\sqrt{3} +2)\cdot(2+\sqrt{3} )+1}{2+\sqrt{3} } =\\\\\\=\frac{3+4\sqrt{3} +4+1}{2+\sqrt{3} } =\frac{8+4\sqrt{3} }{2+\sqrt{3} } =\frac{4\cdot(2+\sqrt{3} )}{2+\sqrt{3} } =4$

Answer 2

Объяснение:

1) ³√(7-√22)•³√(7+√22)=

=³√((7-√22)•(7+√22))=³√(7²-(√22)²)=

=³√(49-22)=³√27=3

2) 1/(4+2√3) + 1/(4-2√3)=

=(4-2√3+4+2√3)/((4+2√3)(4-2√3))=

=8/(4²-(2√3)²)=8/(16-12)=8/4=2

3) 3/(6-2√6) + 3/(6+2√6)=

=(3(6+2√6)+3(6-2√6)) / ((6-2√6)(6+2√6))=

=(18+6√6+18-6√6) / (6²-(2√6)²)=

=36/(36-24)=36/12=3

4) √3+2+1/(2+√3)=

=(√3(2+√3)+2(2+√3)+1)/(2+√3)=

=(2√3+3+4+2√3+1) / (2+√3)=

=(4√3+8) /(2+√3)=4(√3+2)/(2+√3)=4