A)√x-√x+3=1 б)log степень2(1-x)+log степень2(3-x)=3 4степень x+2×2степень x-80=0

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A)

√x - √(x+3) = 1

Let's simplify the equation by squaring both sides to eliminate the square roots:

(√x - √(x+3))^2 = 1^
x - 2√x√(x+3) + x + 3 =
2x + 3 - 2√x√(x+3) =
2x - 2√x√(x+3) = -
2(x - √x(x+3)) = -
2(x - √(x^2 + 3x)) = -2

Solving for x from here involves further simplification.

б)

log base 2(1 - x) + log base 2(3 - x) = 3

Using the rule of logarithms that states log base a(b) + log base a(c) = log base a(b * c), we can combine the logarithms:

log base 2((1 - x)(3 - x)) =
log base 2(3 - 4x + x^2) =
3 - 4x + x^2 = 2^
x^2 - 4x + 3 =
x^2 - 4x - 5 = 0

Solving this quadratic equation will give the value(s) of x.

4^x + 2 * 2^x - 80 = 0

This equation can be simplified by noticing that 4^x = (2^x)^2 and substituting the correct values.

(2^x)^2 + 2 * 2^x - 80 =
Let y = 2^x, then the equation becomes:

y^2 + 2y - 80 =
(y + 10)(y - 8) =
y = -10 or y = 8

Now, solve for x using the values of y:

For y = -10
2^x = -1
This is not valid as 2^x cannot be negative.

For y = 8
2^x =
x = 3

The solution to this equation is x = 3.