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