Ответ:
Объяснение:
2y*(6x+(1/4)*y)²-x*(8x+3y)²=2y*(36x²+3xy+y²/16)-x*(64x²+48xy+9y²)=
=72x²y+6xy²+y³/8-64x³-48x²y-9xy²=-64x³+24x²y-3xy²+y³/8=
=-(64x³-24x²y+3xy²-y³/8)=-((4x)³-3*16x²*(y/2)+3*4x*(y²/4)-(y/2)³)=
=-((4x)³-3*(4x)²*(y/2)+3*4x*(y/2)²-(y/2)³)=-(4x-(y/2))³=((y/2)-4x)³.
I - 0,3х л
II - 5/6 *0,3х
III - 0,3х-26 л
IV - 5/6 *0,3х + 10 л
Всего х л
0,3х+5/6 *0,3х+0,3х-26+5/6 *0,3х + 10=х
0,3х+0,25х+0,3х+0,25х-х=26-10
1,1х-х=16
0,1х=16
<span>х=160</span>
1. 5*(a+c)-b*(a+c)= (5-b)*(a+c)
3. 2*(b^2-6bc+9c^2)=2*(b-3c)^2. Пояснение:(a-b)^2=a^2-2ab+c^2 => (b-3c)^2=(b^2-2*3c+9c^2)
2. (3a-c)*(3a+c)
1)(x-2)(x+2)(2x-1)<0
x=2 x=-2 x=1/2
_ + _ +
---------------------------------------------------
-2 1/2 2
x∈(-∞;-2) U (1/2;2)
2)(3-y)(3+y)(6-5y)≥0
y=3 y=-3 y=1,2
_ + _ +
---------------------------------------------------
-3 1,2 3
y∈[-3;1,2] U [3;∞)
3)<span>(x-1)(x+2)(3x-1)>0
x=1 x=-2 x=1/3
</span> _ + _ +
---------------------------------------------------
-2 1/3 1
x∈(-2;1/3) U (1;∞)
<span>4)(2x-5)(x+0,5)(3x+7)≤0
x=2,5 x=-0,5 x=-7/3
</span>_ + _ +
---------------------------------------------------
-7/3 -0,5 2,5
x∈(-∞;-7/3} U [-0,5;2,5]
X=t²-9t
x²+22x+112=0
D=22²-4*112=36=6²
x₁=(-22-6)/2=-14
x₂=(-22+6)/2=-8
делаем обратную замену
1) t²-9t=-14
t²-9t+14=0
D=81-4*14=81-56=25=5²
t₁=(9+5)/2=7
t₂=(9-5)/2=2
2) t²-9t=-8
t²-9t+8=0
D=81-4*8=81-32=49=7²
t₃=(9+7)/2=8
t₄=(9-7)/2=1
