1)(x-2)(x+2)(2x-1)<0
x=2 x=-2 x=1/2
_ + _ +
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-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]
4x-3y=-1
3x+2y=12
x=(3y-1)/4
x=(12-2y)/3
(3y-1)/4<span>=(12-2y)/3
</span>3(3y-1)=4<span>(12-2y)
</span>9y-3=48-8y
17y=51
y=51/17=3
x(3)=<span>(3</span>·3<span>-1)/4=2</span>
Ответ: точка пресечения прямых — (2;3)
Q=6/3=2
S6=3*(2^6-1)/(2-1)=3*(64-1)=3*63=189
Sn=3*(2^n -1)
(1/3)^x=t
t²+8t-9=0
t1+t2=-8 U t1*t2=-9
t1=-9⇒(1/3)^x=-9 нет решения
t2=1⇒(1/3)^x=1⇒x=0