M4tixx1999viz
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pomocy mam zadanie domowe z matmy daje naj

Pomocy Mam Zadanie Domowe Z Matmy Daje Naj class=

Odpowiedź :

AstraySpirit2308viz

Cześć!

1)

[tex](8+\sqrt8)\cdot8\sqrt8=64\sqrt8+8\sqrt{8^2}=64\sqrt{4\cdot2}+8\cdot8=\\\\=64\sqrt{2^2\cdot2}+64=64\cdot2\sqrt2+64=128\sqrt2+64[/tex]

2)

[tex]\frac{6\sqrt6-3\sqrt3}{2\sqrt2}=\frac{6\sqrt6-3\sqrt3}{2\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{6\sqrt{6\cdot2}-3\sqrt{3\cdot2}}{2\sqrt{2^2}}=\frac{6\sqrt{12}-3\sqrt6}{2\cdot2}=\\\\=\frac{6\sqrt{4\cdot3}-3\sqrt6}{4}=\frac{6\sqrt{2^2\cdot3}-3\sqrt6}{4}=\frac{6\cdot2\sqrt3-3\sqrt6}{4}=\frac{12\sqrt3-3\sqrt6}{4}[/tex]

Catta1eyaviz

[tex](8+\sqrt8)*8\sqrt8=8\sqrt8*8+8\sqrt8*\sqrt8=64\sqrt8+8\sqrt{8*8}=64\sqrt8+64=64\sqrt{4*2}+64=64*2\sqrt2+64=128\sqrt2+64=64(2\sqrt2+1)[/tex]

[tex]\frac{6\sqrt6-3\sqrt3}{2\sqrt2}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2(6\sqrt6-3\sqrt3)}{2*2}=\frac{6\sqrt{12}-3\sqrt{6}}4=\frac{6\sqrt{4*3}-3\sqrt6}4=\frac{12\sqrt3-3\sqrt6}4[/tex]