8*7^(2*x^2-x)-7*8^(2*x^2-x)=0

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вопрос

To solve the equation:

87^(2x^2 - x) - 78^(2x^2 - x) = 0

Let's rewrite this equation using exponential properties:

8(7^2)^(x^2) 7^(-x) - 7(8^2)^(x^2) 8^(-x) = 0
849^(x^2) 7^(-x) - 764^(x^2) 8^(-x) = 0

Now, notice that 49 can be written as 7^2, and 64 can be written as 8^2:

8(7^(x^2))^2 7^(-x) - 7(8^(x^2))^2 8^(-x) = 0

Now, we apply the property (a^m)^n = a^(m*n):

87^(2x^2) 7^(-x) - 78^(2x^2) 8^(-x) = 0
87^(2x^2) 7^(-x) - 78^(2x^2) 8^(-x) = 0

Now, we can simplify this equation further by applying the property a^b * a^c = a^(b+c):

87^(x^2) - 78^(x^2) = 0

Since the bases are different, it is not possible to simplify it further without knowing the value of x.