Задача 72299 вычислить и решить...

61be324ae999dc4d8f52fd53

24.07.2023 15:38:33

вычислить и решить

626fab9a354ab65fb2f43abb

24.07.2023 19:19:12

★

[m]\frac{(\sqrt{19}+\sqrt{2})(\sqrt{38}+\sqrt{57}-\sqrt{6}-2)}{\sqrt{3}+\sqrt{2}}=[/m]
[m]=\frac{\sqrt{19}\sqrt{19 \cdot 2}+\sqrt{2}\sqrt{19 \cdot 2} + \sqrt{19}\sqrt{19 \cdot 3}+\sqrt{2}\sqrt{19 \cdot 3}-\sqrt{19}\sqrt{6}-\sqrt{2}\sqrt{2 \cdot 3}-2\sqrt{19}-2\sqrt{2}}{\sqrt{3}+\sqrt{2}}=[/m]
[m]=\frac{19\sqrt{2}+2\sqrt{19} + 19\sqrt{3}+\sqrt{6}\sqrt{19}-\sqrt{19}\sqrt{6}-2\sqrt{3}-2\sqrt{19}-2\sqrt{2}}{\sqrt{3}+\sqrt{2}}=[/m]
[m]=\frac{(19\sqrt{2}+ 19\sqrt{3})+(\sqrt{6}\sqrt{19}-\sqrt{19}\sqrt{6})+(2\sqrt{19} -2\sqrt{19})+(-2\sqrt{3}-2\sqrt{2})}{\sqrt{3}+\sqrt{2}}=[/m]
[m]=\frac{19(\sqrt{2} + \sqrt{3})-2(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}= \frac{(\sqrt{2} + \sqrt{3})(19-2)}{\sqrt{3}+\sqrt{2}}=19-2=17[/m]

[b]Ответ: 17[/b]

5f3ea7e3faf909182968ddd9

24.07.2023 21:53:43

[m]\sqrt{38}=\sqrt{19\cdot 2}=\sqrt{19}\cdot\sqrt{ 2}[/m]

[m]\sqrt{57}=\sqrt{19\cdot 3}=\sqrt{19}\cdot\sqrt{ 3}[/m]

[m]\sqrt{6}=\sqrt{2\cdot 3}=\sqrt{2}\cdot\sqrt{ 3}[/m]

[m]2=\sqrt{2}\cdot\sqrt{ 2}[/m]


[m]\sqrt{38}+\sqrt{57}-\sqrt{6}-2=\sqrt{19}\cdot(\sqrt{ 2}+\sqrt{3})-\sqrt{2}(\sqrt{ 3}+\sqrt{2})=(\sqrt{ 3}+\sqrt{2})\cdot (\sqrt{19}-\sqrt{2})[/m]


[m]\frac{(\sqrt{19}+\sqrt{2})\cdot(\sqrt{38}+\sqrt{57}-\sqrt{6}-2) }{\sqrt{ 3}+\sqrt{2}}=\frac{(\sqrt{19}+\sqrt{2})\cdot(\sqrt{ 3}+\sqrt{2})\cdot (\sqrt{19}-\sqrt{2}) }{\sqrt{ 3}+\sqrt{2}}=(\sqrt{19}+\sqrt{2})\cdot (\sqrt{19}-\sqrt{2})=(\sqrt{19})^2-(\sqrt{2})^2=19-2=17[/m]

О т в е т. [b]17[/b]