1+log_{2} (4x-5)" alt="log_{2} (x^{2} +3x-4)>1+log_{2} (4x-5)" align="absmiddle" class="latex-formula">
log_{2}2+log_{2} (4x-5)" alt="log_{2} (x^{2} +3x-4)>log_{2}2+log_{2} (4x-5)" align="absmiddle" class="latex-formula">
log_{2}(2*(4x-5))" alt="log_{2} (x^{2} +3x-4)>log_{2}(2*(4x-5))" align="absmiddle" class="latex-formula">
x²+3x-4>8x-10
x²-5x+6>0
x²-5x+6=0
D = (-5)² - 4*6 = 25 - 24 = 1
x∈(-∞;2)∪(3;+∞)
ОДЗ:
0 } \atop {4x-5>0}} \right." alt="\left \{ {{x^{2}+3x-4>0 } \atop {4x-5>0}} \right." align="absmiddle" class="latex-formula">
x²+3x-4=0
D = 3² - 4*(-4) = 9 + 16 = 25 = 5²
x∈(-∞;-4)∪(1;+∞)
4x - 5 = 0
4x = 5
x = 5/4
1,25} \atop {\\\left[\begin{array}{ccc}x1\\\end{array}}} \right." alt="\left \{ {{x>1,25} \atop {\\\left[\begin{array}{ccc}x1\\\end{array}}} \right." align="absmiddle" class="latex-formula">
x∈(1,25;+∞)
Ответ: x∈(1,25;2)∪(3;+∞)