$2)( \frac{c}{c+6} + \frac{36+c^2}{36-c^2} - \frac{c}{c-6} ): \frac{6c+c^2}{(6-c)^2}= \\ ( \frac{c}{c+6} + \frac{(6+c)(6+c)}{(6-c)(6+c)} - \frac{c}{c-6} ): \frac{6c+c^2}{(6-c)^2}= \\ ( \frac{c}{c+6} + \frac{(6+c)}{(6-c)} + \frac{c}{6-c} ): \frac{6c+c^2}{(6-c)^2}=
 \\ ( \frac{6c-c^2+36+c^2+6c+c^2}{(6+c)(6-c)} ): \frac{6c+c^2}{(6-c)^2}= \\ 
( \frac{c^2+12c+36}{(6+c)(6-c)} ): \frac{6c+c^2}{(6-c)^2}= \\ ( \frac{(c+6)^2}{(6+c)(6-c)} ): \frac{6c+c^2}{(6-c)^2}= \\ 
$

$\frac{(c+6)}{(6-c)} : \frac{6c+c^2}{(6-c)^2}= \\ \frac{(c+6)}{(6-c)} * \frac{(6-c)^2}{c(6+c)}= \frac{6-c}{c};$

$3)( \frac{2m}{n+2m}- \frac{4m^2}{n^2+4m^2+4mn} ):( \frac{2m}{n^2-4m^2}+ \frac{1}{2m-n} )= \\ ( \frac{2m}{n+2m}- \frac{4m^2}{(n+2m)^2} ):( \frac{2m}{(n-2m)(n+2m)}- \frac{1}{n-2m})= \\ ( \frac{2mn+4m^2-4m^2}{(n+2m)^2} ):( \frac{2m-n-2m}{(n-2m)(n+2m)} )= \\ \frac{2mn}{(n+2m)^2} *(- \frac{(n-2m)(n+2m)}{n} )= \\ - \frac{2m*(n-2m)}{(n+2m)} = \frac{4m^2-2mn}{(n+2m)} ;$

$4)( \frac{2p}{2p+q}- \frac{4p^2}{4p^2+q^2+4pq} ):( \frac{2p}{4p^2-q^2}+ \frac{1}{q-2p} )= \\ ( \frac{2p}{2p+q}- \frac{4p^2}{(2p+q)^2)} ):( \frac{2p}{(2p-q)(2p+q)}- \frac{1}{2p-q} )= \\ \frac{4p^2+2pq-4p^2}{(2p+q)^2} } : \frac{2p-2p-q}{(2p+q)(2p-q)} = \\ \frac{2pq}{(2p+q)^2} } * (- \frac{(2p-q)(2p+q)}{q} )} = \\ -\frac{2p(2p-q)}{2p+q} = \frac{q-4p^2}{2p+q} ;$

0 0

4 года назад

1)cos(a)-cos(-a), a=π/42)cos(π/2+a)+sin(-a), a=π/3

Инесса Р [889]

1.cosa-cos(-a)=cosa-cosa=0
2.cos(π/2+a)+sin(-a)=-sina-sina=-2sina=-2sinπ/3=-2*√3/2=-√3

0 0

3 года назад

Найдите координаты точек параболы y=-x^2 и y=-9

Мариинна

- x^2 = - 9
x^2 = 9
x = ± 3

x1 = - 3
y1 = - 9

x2 = 3
y2 = - 9

( - 3; - 9) ; ( 3; - 9)

0 0

4 года назад

9x-26=30-5x 5,8-1,6x=0,3-1,8 -4,92y-(0,08 + 5,12)=-0,88y 0,4 (x-9)=0,7+0,3 (x+2)

Янчик2

1)9х-26=30-5х
9х+5х=30+26
14х=56
Х=4
2)5,8-1,6х=0,3-1,8
-1,6х=0,3-1,8-5,8
-1,6х=-7,3
Х=примерно 4,5
3)-4,92у-(0,08+5,12)=-0,88у
-4,92у-5,2=-0,88у
-4,92у+0,88у=5,2
-4,04у=5,2
У=1,28
4)0,4(х-9)=0,7+0,3(х+2)
0,4х-3,6=0,7+0,3х+0,6
0,4х-0,3х=0,7+0,6+3,6
0,1х=4,9
Х=49

0 0

4 года назад

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