Правильный ответ Д. по формуле:а^3+в^3=(а+в)(а^2-ав+в^2)
A)3(1+2xy)(1-2xy)=
=3(1-4x²y²)=3-12x²y²
б)(x²+y³)²=x⁴+2x²y³+y^6
в)(а+b)²-(a-b)²=
=a²+2ab+b²-(a²-2ab+b²)=a²+2ab+b²-
-a²+2ab-b²=4ab
a²,-a²,b²,-b² - это подобные слагаемые
1)9х*2=4
2)х2=4/9
3)х=2/3
ответ 2/3
Lg(5y²-2y+1)/3lg(4y²-5y+1)≤1/3*log(5)7/log(5)7
lg(5y²-2y+1)/lg(4y²-5y+1)≤1
ОДЗ
5y²-2y+1>0
D=4-20=-18<0,a>0⇒y∈(-∞;∞)
4y²-5y+1>0
D=25-16=9
y1=(5-3)/8=1/4
y2=(5+3)/8=1
y<1/4 U y>1
lg(4y²-5y+1)≠0
4y²-5y+1≠1
4y²-5y≠0
y(4y-5)≠0
y≠0 U y≠5/4
y∈(-∞;0) U (0;1/4) U (1;5/4) U (5/4;∞)
log(5y²-2y+1)/lg(4y²-5y+1) -1≤0
[lg(5y²-2y+1)-lg(4y²-5y+1)]/lg(4y²-5y+1)≤0
lg[(5y²-2y+1)/(4y²-5y+1)]/lg(4y²-5y+1)≤0
a){lg[(5y²-2y+1)/(4y²-5y+1)]≥0 (1)
{lg(4y²-5y+1)<0 (2)
(1)(5y²-2y+1)/(4y²-5y+1)≥1
(5y²-2y+1)/(4y²-5y+1) -1≥0
(5y²-2y+1-4y²+5y-1)/(4y²-5y+1)≥0
(y²+3y)/(4y²-5y+1)≥0
y(y+3)/[4(y-1/4)(y-1)≥0
y=0 y=-3 y=1/4 y=1
+ _ + _ +
-----------[-3]------------[0]-----------(1/4)----------(1)----------
y≤-3 U 0 ≤ y<1/4 U y>1
(2)lg(4y²-5y+1)<0
4y²-5y+1<1
4y²-5y<0
y(4y-5)<0
y=0 y=5/4
0<y<5/4
y∈(0;1/4) U (1;5/4)
б){lg[(5y²-2y+1)/(4y²-5y+1)]≤0 (3)
{lg(4y²-5y+1)>0 (4)
(3)-3≤y≤0 U 1/4<y<1
(4)y<0 U y>5/4
y∈[-3;0)
Ответ y∈[-3;0) U (0;1/4) U (1;5/4)
