Question

ПОМОГИТЕ СРОЧНО ! АЛГЕБРА ! ДАЮ 50 БАЛЛОВ !

Answer 1

Ответ:

$1\\1) 5,2+(\sqrt{1,8}) ^2=5,2+1,8=7\\2)9*(-\sqrt{\frac{4}{9} } )^2=9*(-\frac{4}{9}) =-4\\3)15: (\sqrt{\frac{5}{7} } )^2=15*\frac{7}{5} =21\\4)\sqrt{26,01} -(\frac{1}{2} *\sqrt{20} )=5,1-(\frac{1}{2} *2\sqrt{5} )=5,1-\sqrt{5} \\5) (7\sqrt{2})^2 -7*(-\sqrt{2} )^2=98+14=112\\6)(\frac{3}{4} *\sqrt{\frac{16}{27} } )^2+(\frac{2}{5} *\sqrt{\frac{25}{28} } )^2=(\frac{9}{16} *\frac{16}{27}) -(\frac{4}{25} *\frac{25}{28} )=\frac{1}{3} -\frac{1}{7} = \frac{7}{21} -\frac{3}{21} =\frac{4}{21}$

$2\\1)x^{2} =121\\x=б11\\2)x^{2} =-64$

нет решений

$3)x^{2} +13=13\\x^{2} =13-13\\x^{2} =0\\x=0$

$4)29-x^{2} =25\\-29+x^{2} =-25\\x^{2} =-25+29\\x^{2} =4\\x=б2$

$5)(x-5)^2=256\\x^{2} -10x+25=256\\x^{2} -10x+25-256=0\\x^{2} -10x-231=0\\D=(-10)^2-4*(-231)=100+924=1024=32^2\\x_{1} =\frac{10+32}{2} =\frac{42}{2} =21\\x_{2} =\frac{10-32}{2} =-\frac{22}{2} =-11$

$6)(x+\frac{1}{3} )^2=\frac{4}{9} \\\\x^{2} +\frac{2}{3} x+\frac{1}{9} =\frac{4}{9}\\\\x^{2} +\frac{2}{3} x+\frac{1}{9}-\frac{4}{9}=0\\\\x^{2} +\frac{2}{3} x-\frac{3}{9} =0\\\\D=(\frac{2}{3} )^2-4*(-\frac{3}{9} )=\frac{4}{9} +\frac{12}{9} =\frac{16}{9} =(\frac{4}{3} )^2\\\\x_{1} =\frac{-\frac{2}{3}+\frac{4}{3} }{2} =\frac{\frac{2}{3} }{2} =\frac{1}{3} \\\\x_{2} =\frac{-\frac{2}{3}-\frac{4}{3} }{2}=-\frac{\frac{6}{3} }{2} =\frac{3}{3} =1$