Question

44 балла, решите наконец-то уже

Answer 1

а) $3x^{2} -6x-5 = 3 (1+\sqrt{2} )^{2} -6(1+\sqrt{2} )-5= 3( 1^{2} +2*1\sqrt{2} +\sqrt{2} ^{2}) - 6(1+\sqrt{2} )-5=3(1+2\sqrt{2} +2)-6-6\sqrt{2} -5= 3+6\sqrt{2} +6 -6 -6\sqrt{2} -5 = 3-5=-2$

б) $\frac{x^{2} -x-5}{x-1} = \frac{(\sqrt{5} +1)^{2} -(\sqrt{5}+1)-5   }{\sqrt{5}+1-1 } =\frac{(\sqrt{5} ^{2} +2*1\sqrt{5}+1^{2}) -(\sqrt{5}+1)-5    }{\sqrt{5} +1-1} =\frac{5+2\sqrt{5} +1-\sqrt{5}-1 -5}{\sqrt{5} +1-1}= \frac{\sqrt{5} }{\sqrt{5} } =1$

a) $0.3^{-3} +(\frac{3}{7})^{-1} +(0.5)^{-2} *\frac{3}{4} +(-1)^{-8} *6 = (\frac{10}{3} )^{3} +\frac{7}{3} + (\frac{10}{5})^{2}*\frac{3}{4}  + 1 *6 = \frac{1000}{9}  +\frac{7}{3} +\frac{100}{25} *\frac{3}{4}+1*6 = \frac{1000}{9}+\frac{7}{3} +\frac{300}{100}+6 =  \frac{1000}{9}+\frac{7}{3} +\frac{3}{1}+6=\frac{1000+21+27+54}{9} =\frac{1102}{9} =122\frac{4}{9}$

б) $(\frac{2}{3})^{-2}  -(\frac{1}{9})^{-1} +(\frac{6}{17})^{0} }  *\frac{1}{8} -0.25 ^{-2}  *16=(\frac{3}{2})^{2} - \frac{9}{1}  +1*\frac{1}{8} -(\frac{100}{25})^{2}*16 = \frac{9}{4} -9+\frac{1}{8} -\frac{16}{1} *16=\frac{9}{4} -9+\frac{1}{8}- 256= \frac{18-72+1-2048}{8} =-\frac{2101}{8} =-262 \frac{5}{8}$