2 задачи

[smilesims]

[tex]\frac{23\sqrt{9}}{6\sqrt{2} - \sqrt{3} }[/tex][tex]+ \frac{2\sqrt{2} + \sqrt{8}}{\sqrt{18} - \sqrt{2} }[/tex][tex]- \frac{3}{2 }[/tex]
[tex]{(x^{2} - 2x)}^2=[/tex][tex]2(x^{2}-2x) - 3 = 0[/tex]

[mail_dinko]

[tex]\frac{23\sqrt{9}}{6\sqrt{2} - \sqrt{3} } + \frac{2\sqrt{2} + \sqrt{8}}{\sqrt{18} - \sqrt{2} }- \frac{3}{2 }=[/tex]
[tex]=\frac{23.3}{6\sqrt{2} - \sqrt{3} }.\frac{6\sqrt{2}+\sqrt{3} }{6\sqrt{2}+\sqrt{3} }+ \frac{2\sqrt{2} + 2\sqrt{2}}{\sqrt{3.3.2} - \sqrt{2} }- \frac{3}{2 }=[/tex]
[tex]=\frac{69(6\sqrt{2}+\sqrt{3} )}{(6\sqrt{2})^2 - (\sqrt{3})^2 }+ \frac{4\sqrt{2}}{3\sqrt{2} - \sqrt{2} }- \frac{3}{2 }=[/tex]
[tex]=\frac{69(6\sqrt{2}+\sqrt{3} )}{72-3}+ \frac{4\sqrt{2}}{2\sqrt{2} }- \frac{3}{2 }=[/tex]
[tex]=\frac{69(6\sqrt{2}+\sqrt{3} )}{69}+ 2- \frac{3}{2 }=[/tex]
[tex]=6\sqrt{2}+\sqrt{3}+\frac{4}{2 }-\frac{3}{2 }=[/tex]
[tex]=6\sqrt{2}+\sqrt{3}+\frac{1}{2 }[/tex]

[mail_dinko]

[tex]{(x^{2} - 2x)}^2=2(x^{2}-2x) - 3 = 0[/tex]
Полагаме
[tex](x^{2} - 2x)=t[/tex]
[tex]t^2=2t-3=0[/tex]
[tex]t^2-2t+3=0[/tex]
[tex]D=1-3<0[/tex]
няма решение