4 sqrt x =1/3
sqrt x = 1\12
x= 1\144
P.S sqrt - корень
![1) 3^{x}\ \textgreater \ \frac{1}{3^5} \\ \\ 3^x\ \textgreater \ 3^{-5} \\ \\ x\ \textgreater \ -5](https://tex.z-dn.net/?f=1%29++3%5E%7Bx%7D%5C+%5Ctextgreater+%5C++%5Cfrac%7B1%7D%7B3%5E5%7D+%5C%5C++%5C%5C+%0A3%5Ex%5C+%5Ctextgreater+%5C++3%5E%7B-5%7D+%5C%5C++%5C%5C+x%5C+%5Ctextgreater+%5C+-5)
x∈(-5; +∞)
![2)( \frac{5}{7}) ^{3x+4} \geq ( \frac{5}{7})^{2}](https://tex.z-dn.net/?f=2%29%28+%5Cfrac%7B5%7D%7B7%7D%29+%5E%7B3x%2B4%7D+++%5Cgeq+%28+%5Cfrac%7B5%7D%7B7%7D%29%5E%7B2%7D++)
Поскольку 5/7<1, то пр одинаковых основаниях степени
3x+4≤2
3x≤2-4
3x≤-2
x≤-2/3
x∈(-∞; -2/3]
3) 4²ˣ - 5*4ˣ+4<0
Пусть 4ˣ=y, тогда 4²ˣ=у²
у²-5у+4<0
y²-4y -y+4<0
y(y-4)-(y-4)<0
(y-4)(y-1)<0
y∈(1; 4)
Проведем обратную замену
1<4ˣ<4
4⁰<4ˣ<4¹
0<x<1
x∈(0; 1)
(3х - 5у) (3х + 5у); б) (аb^3)^2 - (4x^2)^2 = (ab^3 - 4x^2) (ab^3 + 4x^2
1) (x+5)*(x+5)=(x-5)*(x-5)
x^2+10x+25=x^2-10x+25
x^2+10x-x^2+10x=25-25
20x=0
x=0
2) (4x+1)*x=(3x-1)*3
4x^2+x=9x-3
4x^2+x-9x+3=0
4x^2-8x+3=0
D= (-8)^2-4*4*3=64-48=16
x1= (8-4)/2*4=4/8=1/2=0,5
x2=(8+4)/2*4=12/8=3/2= 1,5
3) (x-4)*x=(10+2x)*2
x^2-4x=20+4x
x^2-4x-20-4x=0
x^2-8x-20=0
x1+x2=8 x1= 10
x1*x2= -20 x2= -2
4) (4x+1)*(x+1)= (3x-8)(x-3)
4x^2+4x+x+1=3x^2-9x-8x+24
4x^2+4x+x+1-3x^2+9x+8x-24=0
x^2+22x-23=0
x1+x2= -22 x1= -23
x1*x2= -23 x2= 1