Matematyka
wikusia24012000
2017-06-24 11:15:49
Proszę o rozwiązanie zadania. Z góry dziękuję
Odpowiedź
perelka2945
2017-06-24 13:11:22

Wykorzystamy wzory skróconego mnożenia: [latex](apm b)^2=a^2pm 2ab+b^2\\(a-b)(a+b)=a^2-b^2[/latex] [latex]a)\left(sqrt{4-sqrt{15}}-sqrt{4+sqrt{15}} ight)^2\\=(sqrt{4-sqrt{15}})^2-2sqrt{4-sqrt{15}}cdotsqrt{4+sqrt{15}}+(sqrt{4+sqrt{15}})^2\\=4-sqrt{15}-2sqrt{(4-sqrt{15})(4+sqrt{15})}+4+sqrt{15}\\=8-2sqrt{4^2-(sqrt{15})^2}\\=8-2sqrt{16-15}=8-2sqrt1=8-2=6[/latex] [latex]b)\sqrt{2(2-2sqrt2)^2}+sqrt{2(2+2sqrt2)^2}\\=sqrt{2left(2^2-2cdot2cdot2sqrt2+(2sqrt2)^2 ight)}+sqrt{2left(2^2+2cdot2cdot2sqrt2+(2sqrt2)^2 ight)}\\=sqrt{2(4-8sqrt2+8)}+sqrt{2(4+8sqrt2+8)}\\=sqrt{2(12-8sqrt2)}+sqrt{2(12+8sqrt2)}\\=sqrt{24-16sqrt2}+sqrt{24+16sqrt2}\\=sqrt{16+8-16sqrt2}+sqrt{16+8+16sqrt2}\\=sqrt{4^2-2cdot4cdot2sqrt2+(2sqrt2)^2}+sqrt{4^2+2cdot4cdot2sqrt2+(2sqrt2)^2}\\=sqrt{(4-2sqrt2)^2}+sqrt{(4+2sqrt2)^2}\\=|4-2sqrt2|+|4+2sqrt2|\\=4-2sqrt2+4+2sqrt2=8[/latex] [latex]c)\left[(2sqrt2-sqrt7)^frac{1}{2}+(2sqrt2+sqrt7)^frac{1}{2} ight]^2\\=left[(2sqrt2-sqrt7)^frac{1}{2} ight]^2+2cdot(2sqrt2-sqrt7)^frac{1}{2}cdot(2sqrt2+sqrt7)^frac{1}{2}+left[(2sqrt2+sqrt7)^frac{1}{2} ight]^2\\=2sqrt2-sqrt7+2left[(2sqrt2-sqrt7)(2sqrt2+sqrt7) ight]^frac{1}{2}+2sqrt2+sqrt7\\=4sqrt2+2left[(2sqrt2)^2-(sqrt7)^2 ight]^frac{1}{2}=4sqrt2+2(8-7)^frac{1}{2}\\=4sqrt2+2cdot1^frac{1}{2}=4sqrt2+2[/latex] [latex]d)\sqrt{11-6sqrt2}+sqrt{11+6sqrt2}\\=sqrt{9+2-6sqrt2}+sqrt{9+2+6sqrt2}\\=sqrt{3^2-2cdot3cdotsqrt2+(sqrt2)^2}+sqrt{3^2+2cdot3cdotsqrt2+(sqrt2)^2}\\=sqrt{(3-sqrt2)^2}+sqrt{(3+sqrt2)^2}=|3-sqrt2|+|3+sqrt2|\\=3-sqrt2+3+sqrt2=6[/latex]

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