Tìm x biết rằng

Let's simplify the equation step by step:

(2^x+1)^2 + 3×(2^2+1) = 2^2×10

First, let's simplify the expression inside the parentheses:
(2^x+1)^2 = (2^x)^2 + 2(2^x)(1) + 1
= 2^(2x) + 2^(x+1) + 1

Now let's simplify the expression inside the second set of parentheses:
3×(2^2+1) = 3×(4+1)
= 3×5
= 15

Now let's substitute these simplified expressions back into the original equation:
2^(2x) + 2^(x+1) + 1 + 15 = 2^2×10

Simplifying further:
2^(2x) + 2^(x+1) + 16 = 4×10
2^(2x) + 2^(x+1) + 16 = 40

Now let's simplify the exponents:
2^(2x) = (2^x)^2
2^(x+1) = 2^x × 2^1 = 2^x × 2

Substituting these back into the equation:
(2^x)^2 + 2^x × 2 + 16 = 40

Simplifying further:
(2^x)^2 + 2^x × 2 + 16 - 40 = 0
(2^x)^2 + 2^x × 2 - 24 = 0

Now we have a quadratic equation in terms of (2^x):
Let's factorize it:
(2^x + 6)(2^x - 4) = 0

Setting each factor equal to zero:
2^x + 6 = 0 or 2^x - 4 = 0

For the first equation, 2^x cannot be negative, so there are no solutions.

For the second equation:
2^x = 4
Taking the logarithm base 2 of both sides:
x = log2(4)
x = 2

Therefore, the solution to the equation is x = 2.