1
a)log(1/4)64=-3
b)5^(2log(5)7)=5^(log(5)49=49
c)log(2)192-log(2)3=log(2)(192/2)=log(2)64=6
2
{2x+3>0⇒2x>-3⇒x>-1,5
{2x+3=64⇒2x=64-3⇒2x=61⇒x=61:2⇒x=30,5
Ответ 30,5
3
{x+3>0⇒x>-6
{x+6<5 (основание меньше 1,знак меняется)⇒x<5-6⇒x<-1
x∈(-6;-1)
4
a)x>0
log(2)x=t
t²-3t-4=0
t1+t2=3 U t1*t2=-4
t1=-1⇒log(2)x=-1⇒x=1/2
t2=4⇒log(2)x=4⇒x=16
b)ОДЗ
{x+2>0⇒x>-2
{x>0
x∈(0;∞)
log(3)[x(x+2)]=1
x²+2x=3
x²+2x-3=0
x1+x2=-2 U x1*x2=-3
x1=-3∉ОДЗ
x2=1
5
a)ОДЗ
{x-3>0⇒x>3
{x-2>0⇒x>2
x∈(3;∞)
log(2)[(x-3)(x-2)]≥1
(x-3)(x-2)≥2
x²-3x-2x+6-2≥0
x²-5x+4≥0
x1+x2=5 U x1*x2=4
x1=1 U x2=4
+ _ +
------[1]--------(3)-------[4]-------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[4;∞)
b)
ОДЗ
x²-4>0⇒(x-2)(x+2)>0⇒x<-2 U x>2
x∈(-∞;-2) U (2;∞)
log(1/3)(x²-4)/log(3)5≥0
log(3)(x²-4)/log(3)5≤0
log(5)(x²-4)≤0
x²-4≤1
(x²-5≤0
(x-√5)(x+√5)≤0
x=√5,x=-√5
//////////////////////////////////////////////
-------[-√5]------(-2)--------(2)---------[√5]--------
////////////////////////// \\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[-√5;-2) U (2;√5]
1) (а+в+с)(а+в+с)=
а^2+ав+ас+ав+в^2+вс+ас+вс+с^2=
а^2+в^2+с^2+2ав+2вс+2ас=
а^2+в^2+с^2+2(ав+вс+ас);
значит
9*9= а^2+в^2+с^2+2*(-10);
а^2+в^2+с^2= 81+20= 101;
2) х(х^2-9)+6(х^2-9)=0;
(х^2-9)(х+6)=0;
(х-3)(х+3)(х+6)=0;
х1=3; х2=-3; х3=-6;
3) (16х^2-49)=0;
(4х-7)(4х+7)=0;
х1=7/4=1,75; х2=-7/4=-1,75;
4) 85^2-15^2= (85-15)(85+15)=
70*100= 7000;
5) (х-1)^2=9; извлекаем корень;
х-1=+-3;
х1=1+3=4; х2=1-3=-2;
