1)4^2x-3 -3*4^x-2 -1=0 2)5^2x+1 +6^x+1>30+5^x30^x SOS SOS SOS SOS SOS

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вопрос

1) Let's solve the equation 4^(2x-3) - 3*4^(x-2) - 1 = 0

First, we can rewrite the equation using exponent properties:
(4^2)^x 4^(-3) - 3 (4^1)^x 4^(-2) - 1 = 0
16^x 1/64 - 3 4^x 1/16 - 1 = 0

Simplify further:
(16^x)/64 - (34^x)/16 - 1 = 0
(1/4)(16^x) - (3/4)(4^x) - 1 = 0
(1/4)(2^x)^4 - (3/4)*(2^x)^2 - 1 = 0

Now, let y = 2^x. The equation becomes:
(1/4)y^4 - (3/4)y^2 - 1 = 0

This is a quadratic equation in terms of y^2:
(1/4)z^2 - (3/4)z - 1 = 0

Solve this quadratic equation for z and then substitute back in y = 2^x to find the values of x that satisfy the original equation.

2) To solve the inequality 5^(2x+1) + 6^(x+1) > 30 + 5^x:

We can rewrite the inequality as:
255^2x + 66^x > 30 + 5^x

Now, let y = 5^x. The inequality becomes:
25y^2 + 6y > 30 + y

Rearranging terms gives:
25y^2 + 6y - y > 30
25y^2 + 5y - 30 > 0
5y(5y + 1) - 30 > 0
25y^2 + 5y - 30 > 0

Now solve the quadratic inequality above to find the range of values for y (and hence, x) that satisfy the original inequality.