√11*11^2x=1/11
11^1/2*11^2x=11^(-1)
11(1/2+2x)=11^(-1)
1/2+2x=-1
2x=-1-1/2
x=-3/4
В первом ящику х кг
во втором x+12
в третьем х+12+5=х+17
x+x+12+x+17=95
3x+29=95
3x=66
x=22
Log(5)(2-x)+0,5log(5)(4x-11)=0
{2-x>0⇒x<2
{4x-11≠0⇒x≠2,75
x∈(-∞;2)
log(5)(2-x)+log(5)√(4x-11)²=0
log(5)(2-x)+log(5)|4x-11|=0
log(5)[(2-x)*|4x-11|]=0
(2-x)*|4x-11|=1
x∈(-∞;2)⇒|4x-11|=11-4x
(2-x)(11-4x)=0
x=2не удов усл
х=2,75 не удов усл
Ответ нет решения
lgx²+lg(x+4)²≥-lg1/9
{x≠0
{x≠-4
x∈(-∞;-4) U (-4;0) U (0;∞)
lg[x²(x+4)²]≥lg9
x²(x+4)²≥9
x²(x+4)²-9≥0
(x(x+4)-3)(x(x+4)+3)≥0
(x²+4x-3)(x²+4x+3)≥0
x²+4x-3=0
D=16+12=28
x1=(-4-2√7)/2=-2-√7 U x2=-2+√7
x²+4x+3=0
x1+x2=-4 U x1*x2=3⇒x1=-3 U x2=-1
+ _ + _ +
-------------[-2-√7]------(-4)-------[-3]------------[-1]------(0)--------[-2+√7]--------------
////////////////////////////////////\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\/////////////////////////////////
x∈(-∞;-2-√7] U [-3;-1] U [-2+√7;∞)
1
1+cosx+cos(x/2)=0
2cos²(x/2)+cos(x/2)=0
cos(x/2)*(2cos(x/2)+1)=0
cos(x/2)=0⇒x/2=π/2+πk⇒x=π+2πk,k∈z
cos(x/2)=-1/2
x/2=+-2π/3+2πn,n∈z⇒x=+-4π/3+4πn,n∈z
2
2cos²x-3sinx-2=0
2(cos²x-1)-3sinx=0
-2sin²x-3sinx=0
-sinx(2sinx+3)=0
sinx=0⇒x=πn,n∈z
sinx=-1,5<-1 нет решения
3
сos2x-(cosx+cos3x)=0
cos2x-2cos2xcosx=0
cos2x(1-2cosx)=0
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z
cosx=1/2⇒x=+-π/3+2πk,k∈z
X=2
y=5
$$3"=:/#4:=#="="=#÷