Решите заданьице, позязя. Created by Twilighte007. algebra-ru

Решите заданьице, позязя...

kmike21

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$\frac{15a^2-10ab}{3ab-2b^2} = \frac{5a (3a-2b)}{b (3a-2b)}=\frac{5a}{b }=\frac{5(-2)}{-0,1}=100$

$\frac{9c^2-4d^2}{18c^2d-12cd^2} =\frac{(3c-2d)(3c+2d)}{6cd (3c-2d)} =\frac{3c+2d}{6cd}=\frac{3 *\frac{2}{3} +2* \frac{1}{2} }{6* \frac{2}{3} * \frac{1}{2} }= \frac{2+1}{2} = \frac{3}{2}=1 \frac{1}{2}$

$\frac{6x^2+12xy}{5xy+10y^2} = \frac{6x (x+2y)}{5y (x+2y)} = \frac{6x}{5y} =\frac{6* \frac{2}{3} }{5(-0,4)} = \frac{4}{-2} =-2$

$\frac{x^2+6xy+9y^2}{4x^2+12xy} = \frac{(x+3y)^2}{4x (x+3y)} = \frac{x+3y}{4x} = \frac{-0,2+3(-0,6)}{4(-0,2)}=\frac{-0,2-1,8}{-0,8}= \frac{-2}{-0,8} = \frac{20}{8}= \\ =\frac{5}{2} = 2,5$

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sedinalana 1
5a(3a-1b)/[b(3a-2b)]=5a/b
a=-2;b=-0,1 -2*5/(-0,1)=100
2
(3c-2d)(3c+2d)/[6cd(3c-2d)]=(3c+2d)/(6cd)
c=2/3;d=1/2 (3*1/3+2*1/2)/(6*2/3*1/2)=(2+1)/2=3/2=1,5

3
6x(x+2y)/[5y(x+2y)]=6x/(5y)
x=2/3;y=-0,4 6*2/3/(-0,4*5)=4/(-2)=-2
4
(x+3y)²/[4x(x+3y)]=(x+3y)/(4x)
x=-0,2;y=-0,6 (-0,2-0,6*3)/(-0,2*4)=(-0,2-1,8)/(-0,8)=-2/(-0,8)=0,25

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