Question

Сделайте номер 17 и 20 с решениеми за старания 20 баллов .

Answer 1

17.

(√6 + √5)² - √120=6+2√30+5-√120=11+2√30 - √(4*30)=11+ 2√30 - 2√30=11

$\tt\displaystyle(\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3} })^{2}=2+\sqrt{3}  +2*\sqrt{2+\sqrt{3} }*\sqrt{2-\sqrt{3} } +2-\sqrt{3}   }=\\  \\ 4+2*(2-\sqrt{3})=8-2\sqrt{3}$

√60+(√5 - √3)²=√60+5-2√15+3=√(4*15) - 2√15 + 8=2√15 - 2√15 + 8=8

$\tt\displaystyle(\sqrt{3+2\sqrt{2} }-\sqrt{3-2\sqrt{2}})^{2}=3+2\sqrt{2} -2(3+2\sqrt{2})(3-2\sqrt{2})+3-2\sqrt{2} =6-2(9-4*2)=6-2=4$

18.  

х=√5 + 2

х² - 4х - 5=(√5 +2)² - 4(√5 +2) - 5=5+4√5+4 - 4√5 - 8 - 5= - 4

.

у=√3 - 1

у²+2у+3=(√3 - 1)²+2(√3 - 1)+3=3 - 2√3 + 1+2√3 - 2+3=5

19.

х² - 3=(х-√3)(х+√3)

4у² - 5=(2у-√5)(2у+√5)

2-с²=(√2 - с)(√2+с)

4-а=(2+√а)(2-√а)

20.

$\frac{\sqrt{6}-\sqrt{2}}{\sqrt{8} }=\frac{\sqrt{2}(\sqrt{3} -1) }{\sqrt{2}*\sqrt{4}  } =\frac{\sqrt{3} -1}{2} \\ \\ \\ \frac{\sqrt{6}+\sqrt{15}  }{\sqrt{8}+\sqrt{20}  } =\frac{\sqrt{3}(\sqrt{2} +\sqrt{5} ) }{\sqrt{4}(\sqrt{2} +\sqrt{5} ) } =\frac{\sqrt{3} }{2} \\ \\ \\ \frac{10-2\sqrt{5} }{\sqrt{20}-2 } =\frac{10-2\sqrt{5} }{\sqrt{4*5}-2 } =\frac{2(5-\sqrt{5}) }{2\sqrt{5}-2 } =\frac{2(5-\sqrt{5})}{2(\sqrt{5}-1) } =\frac{5-\sqrt{5} }{\sqrt{5} -1}$

$\frac{4+2\sqrt{2} }{\sqrt{8}+2 } =\frac{2(2+\sqrt{2}) }{2\sqrt{2}+2 } =\frac{2(\sqrt{2}+2) }{2(\sqrt{2}+1) } =\frac{\sqrt{2}+2 }{\sqrt{2}+1 }$