Ответы
Ответ дал:
0
в квадрате, я полагаю? тогда лови: ![(3sqrt{2}-3)^2=27-18sqrt{2}; (3sqrt{2}-3)^2=27-18sqrt{2};](https://tex.z-dn.net/?f=%283sqrt%7B2%7D-3%29%5E2%3D27-18sqrt%7B2%7D%3B)
на множители разложить двучлен? да пожалуйста:![3sqrt{2}-3=3(sqrt{2}-1) 3sqrt{2}-3=3(sqrt{2}-1)](https://tex.z-dn.net/?f=3sqrt%7B2%7D-3%3D3%28sqrt%7B2%7D-1%29)
и ![B=frac{2}{3}sqrt{12} B=frac{2}{3}sqrt{12}](https://tex.z-dn.net/?f=B%3Dfrac%7B2%7D%7B3%7Dsqrt%7B12%7D)
и ![B=sqrt{frac{4}{9}*12} B=sqrt{frac{4}{9}*12}](https://tex.z-dn.net/?f=B%3Dsqrt%7Bfrac%7B4%7D%7B9%7D%2A12%7D)
и ![B=sqrt{5frac{1}{3}} B=sqrt{5frac{1}{3}}](https://tex.z-dn.net/?f=B%3Dsqrt%7B5frac%7B1%7D%7B3%7D%7D)
, значит ![A textgreater B A textgreater B](https://tex.z-dn.net/?f=A+textgreater++B)
![frac{16-6}{sqrt{2}-4}=frac{10}{sqrt{2}-4}=frac{10(sqrt{2}+4)}{(sqrt{2}-4)(sqrt{2}+4)}=frac{10(sqrt{2}+4)}{2-16}=-frac{5}{7}(sqrt{2}+4) frac{16-6}{sqrt{2}-4}=frac{10}{sqrt{2}-4}=frac{10(sqrt{2}+4)}{(sqrt{2}-4)(sqrt{2}+4)}=frac{10(sqrt{2}+4)}{2-16}=-frac{5}{7}(sqrt{2}+4)](https://tex.z-dn.net/?f=frac%7B16-6%7D%7Bsqrt%7B2%7D-4%7D%3Dfrac%7B10%7D%7Bsqrt%7B2%7D-4%7D%3Dfrac%7B10%28sqrt%7B2%7D%2B4%29%7D%7B%28sqrt%7B2%7D-4%29%28sqrt%7B2%7D%2B4%29%7D%3Dfrac%7B10%28sqrt%7B2%7D%2B4%29%7D%7B2-16%7D%3D-frac%7B5%7D%7B7%7D%28sqrt%7B2%7D%2B4%29)
![frac{2sqrt{5}}{sqrt{5}*sqrt{3}}=frac{2}{sqrt{3}}=frac{2}{3}sqrt{3} frac{2sqrt{5}}{sqrt{5}*sqrt{3}}=frac{2}{sqrt{3}}=frac{2}{3}sqrt{3}](https://tex.z-dn.net/?f=frac%7B2sqrt%7B5%7D%7D%7Bsqrt%7B5%7D%2Asqrt%7B3%7D%7D%3Dfrac%7B2%7D%7Bsqrt%7B3%7D%7D%3Dfrac%7B2%7D%7B3%7Dsqrt%7B3%7D)
![frac{1}{1+3sqrt{5}}+frac{1}{1-3sqrt{5}}=frac{1-3sqrt{5}}{(1-3sqrt{5})(1+3sqrt{5})}+frac{1+3sqrt{5}}{(1-3sqrt{5})(1+3sqrt{5})}=frac{1-3sqrt{5}+1+3sqrt{5}}{1^2-(3sqrt{5})^2}=\frac{2}{1-45}=frac{2}{-44}=-frac{1}{22} frac{1}{1+3sqrt{5}}+frac{1}{1-3sqrt{5}}=frac{1-3sqrt{5}}{(1-3sqrt{5})(1+3sqrt{5})}+frac{1+3sqrt{5}}{(1-3sqrt{5})(1+3sqrt{5})}=frac{1-3sqrt{5}+1+3sqrt{5}}{1^2-(3sqrt{5})^2}=\frac{2}{1-45}=frac{2}{-44}=-frac{1}{22}](https://tex.z-dn.net/?f=frac%7B1%7D%7B1%2B3sqrt%7B5%7D%7D%2Bfrac%7B1%7D%7B1-3sqrt%7B5%7D%7D%3Dfrac%7B1-3sqrt%7B5%7D%7D%7B%281-3sqrt%7B5%7D%29%281%2B3sqrt%7B5%7D%29%7D%2Bfrac%7B1%2B3sqrt%7B5%7D%7D%7B%281-3sqrt%7B5%7D%29%281%2B3sqrt%7B5%7D%29%7D%3Dfrac%7B1-3sqrt%7B5%7D%2B1%2B3sqrt%7B5%7D%7D%7B1%5E2-%283sqrt%7B5%7D%29%5E2%7D%3D%5Cfrac%7B2%7D%7B1-45%7D%3Dfrac%7B2%7D%7B-44%7D%3D-frac%7B1%7D%7B22%7D)
на множители разложить двучлен? да пожалуйста:
Вас заинтересует
1 год назад
1 год назад
6 лет назад
6 лет назад
8 лет назад
9 лет назад