\( 2 \cdot 5 ^ { x } - 2 ^ { x - 1 } - \sqrt { 7 \cdot 10 ^ { x } - 2 \cdot 4 ^ { x } - 5 \cdot 25 ^ { x } } > 0 \)

To tackle the inequality, let's first simplify the expression inside the square root. Note that \(10^x = 2^x \cdot 5^x\) and \(25^x = (5^2)^x = 5^{2x}\). With this in mind, we rewrite \(7 \cdot 10^x - 2 \cdot 4^x - 5 \cdot 25^x\): \[ 7 \cdot (2^x \cdot 5^x) - 2 \cdot (2^x)^2 - 5 \cdot (5^x)^2 \] This gives us: \[ 7 \cdot 2^x \cdot 5^x - 2 \cdot 2^{2x} - 5 \cdot 5^{2x}. \] Next, we recognize \(4^x = (2^2)^x = 2^{2x}\) and express everything together as: \[ 7 \cdot 2^x \cdot 5^x - 2^{2x+1} - 5 \cdot 5^{2x}. \] This certainly complicates the analysis, so the key here is to find suitable values of \(x\) that keep the entire expression positive. To find boundaries or possible solutions, it can be helpful to check some convenient values for \(x\) such as \(x=0\) and \(x=1\): 1. For \(x = 0\): \[ 2 \cdot 5^0 - 2^{-1} - \sqrt{7 \cdot 10^0 - 2 \cdot 4^0 - 5 \cdot 25^0} = 2 - 0.5 - \sqrt{7 - 2 - 5} = 2 - 0.5 - \sqrt{0} = 1.5 > 0 \] 2. For \(x = 1\): \[ 2 \cdot 5^1 - 2^{1-1} - \sqrt{7 \cdot 10^1 - 2 \cdot 4^1 - 5 \cdot 25^1} = 10 - 1 - \sqrt{70 - 8 - 125} = 10 - 1 - \sqrt{-63} \] which is not applicable as it goes under the square root, indicating negativity. Finding roots analytically or numerically for \(x\) or using graphing technology might clarify where the expression turns negative. In conclusion, careful substitution will show where the inequality holds firmly. A detailed check for values between dominantly positive and negative outcomes might show solutions regarding the real number sets, but numerical approximation or plotting will truly reveal their nature!