Question

Найти числовое значение выражения: 1) xy/x^{2} +xy+ y^{2}, если а)x= -3 y= -2 б)x= -0,4 y=0,5 в)x= 1/3 y= -3/5 г)x= -1/3 y= -0,25 2) а-в/а^{2}-ав+2в^{2},если а) а= -4; в = -1 б) а = 3/4; в= -0,5 в) а =-2/3; в=0,5 г) а = 1 1/3; в = -1,5

Answer 1

Решение на фото                                                                                  

Answer 2

При х=-3; y=-2:
$\frac{xy}{x^2+xy+y^2}=\frac{xy(x-y)}{(x^2+xy+y^2)(x-y)}=\\\\ \frac{xy(x-y)}{x^3-y^3}=\frac{-3*(-2)*(-3-(-2))}{(-3)^3-(-2)^3}=\frac{6*(-1)}{-27+8}=\frac{6}{19}$

при х=-0.4, y=0.5
$=\frac{xy(x-y)}{x^3-y^3}=\frac{-0.4*0.5*(-0.4-0.5)}{(-0.4)^3-0.5^3}=\\\\ \frac{-0.2*(-0.9)}{-0.064-0.125}=\frac{0.18}{-0.189}=\frac{180}{189}=\frac{20}{21}$

при x=1/3 y=-3/5
$=\frac{xy(x-y)}{x^3-y^3}=\frac{\frac{1}{3}*\frac{-3}{5}*(\frac{1}{3}-\frac{-3}{5})}{(\frac{1}{3})^3-(\frac{-3}{5})^3}=\\\\ \frac{\frac{-1}{5}*\frac{5+9}{15}}{\frac{1}{27}+\frac{27}{125}}=\\\\ \frac{(-1*14)*125*27}{5*15*(125+27*27)}=\frac{-14*5*9}{125+729}=-\frac{14*5*9}{854}=\frac{45}{61}$

при x=-1/3 y=-0.25=-1/4
$=\frac{xy(x-y)}{x^3-y^3}=\frac{\frac{-1}{3}*\frac{-1}{4}*(\frac{-1}{3}-\frac{-1}{4})}{(-\frac{1}{3})^3-(\frac{-1}{4})^3}=\frac{\frac{1}{12}*\frac{-4+3}{12}}{\frac{-1}{27}+\frac{1}{64}}=\\\\ \frac{-\frac{1}{12*12}}{\frac{-64+27}{64*27}}=\frac{64*27}{12*12*37}=\frac{4*3}{37}=\frac{12}{37}$

2. если a=-4 b=-1:
$\frac{a-b}{a^2-ab+2b^2}=\frac{-4-(-1)}{(-4)^2-(-4)*(-1)+2*(-1)^2}=\frac{-4+1}{16-4+2}=\frac{-3}{14}=-\frac{3}{14}$

при а=3/4 b=-0.5=-1/2
$=\frac{\frac{3}{4}-\frac{-1}{2}}{(\frac{3}{4})^2-\frac{3}{4}*\frac{-1}{2}+2*(-\frac{1}{2})^2}=\\\\ \frac{\frac{3+2}{4}}{\frac{9+6+8}{16}}=\frac{5*16}{4*23}=\frac{5*4}{23}=\frac{20}{23}$

при a=-2/3 b=0.5=1/2
$=\frac{\frac{-2}{3}-\frac{1}{2}}{(\frac{-2}{3})^2-\frac{-2}{3}*\frac{1}{2}+2*(\frac{1}{2})^2}=\\\\ \frac{\frac{-4-3}{6}}{\frac{4}{9}+\frac{1}{3}+\frac{1}{2}}=\frac{-7*9*2}{6*(8+6+9)}=\frac{-7*3}{23}=-\frac{21}{37}=$

при a=1 1/3=4/3  b=-1.5=-3/2
$=\frac{\frac{4}{3}-\frac{-3}{2}}{(\frac{4}{3})^2-\frac{4}{3}*\frac{-3}{2}+2*(-\frac{3}{2})^2}=\\\\ \frac{\frac{8+9}{6}}{\frac{16}{9}+2+\frac{9}{2}}=\frac{17*9*2}{6*(32+36+81)}=\frac{17*3}{149}=\frac{51}{149}$