Ответы на вопрос:
ответ:
пошаговое объяснение:
х+4 4/19 = 6 2/19
х=6 2/19-4 4/19=2-2/19
x=1 17/19
1 17/19+4 4/19=6 2/19
6 2/19=6 2/19
0\; \; ,\; \; \; (3x-1)^2>0\; \; \Rightarrow \; \; 3x-1\ne 0\; ,\\\\3x\ne 1\; \; ,\; \; x\ne \frac{1}{3}\\\\x\in (-\infty ;\frac{1}{3}\, )\cup (\, \frac{1}{3}\, ;\, +\infty )" class="latex-formula" id="TexFormula2" src="https://tex.z-dn.net/?f=4%29%5C%3B%20%5C%3B%209x%5E2-6x%2B1%3E0%5C%3B%20%5C%3B%20%2C%5C%3B%20%5C%3B%20%5C%3B%20%283x-1%29%5E2%3E0%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%203x-1%5Cne%200%5C%3B%20%2C%5C%5C%5C%5C3x%5Cne%201%5C%3B%20%5C%3B%20%2C%5C%3B%20%5C%3B%20x%5Cne%20%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5Cx%5Cin%20%28-%5Cinfty%20%3B%5Cfrac%7B1%7D%7B3%7D%5C%2C%20%29%5Ccup%20%28%5C%2C%20%5Cfrac%7B1%7D%7B3%7D%5C%2C%20%3B%5C%2C%20%2B%5Cinfty%20%29" title="4)\; \; 9x^2-6x+1>0\; \; ,\; \; \; (3x-1)^2>0\; \; \Rightarrow \; \; 3x-1\ne 0\; ,\\\\3x\ne 1\; \; ,\; \; x\ne \frac{1}{3}\\\\x\in (-\infty ;\frac{1}{3}\, )\cup (\, \frac{1}{3}\, ;\, +\infty )">
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