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July 2022 1 5 Report

Решите полностью два примера можно фотография 1) 2x-1/x+7 меньше 3x+4/x-1 2)8-3x/x+1 меньше или равно 4x+1/3-x

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NNNLLL54

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$1)\; \; \frac{2x-1}{x+7} \ \textless \ \frac{3x+4}{x-1} \; ,\; \; \; ODZ:\; \; x\ne -7\; ,\; \; x\ne 1\\\\ \frac{2x-1}{x+7} -\frac{3x+4}{x-1} \ \textless \ 0\\\\ \frac{(2x-1)(x-1)-(3x+4)(x+7)}{(x+7)(x-1)} \ \textless \ 0\\\\ \frac{2x^2-3x+1-(3x^2+25x+28)}{(x+7)(x-1)} \ \textless \ 0\\\\ \frac{-x^2-28x-27}{(x+7)(x-1)} \ \textless \ 0\; ,\; \; \; \frac{x^2+28x+27}{(x+7)(x-1)} \ \textgreater \ 0\; ,\; \; \; \frac{(x+1)(x+27)}{(x+7)(x-1)} \ \textgreater \ 0\\\\+++(-27)---(-7)+++(-1)---(1)+++\\\\x\in (-\infty ,-27)\cup (-7,-1)\cup (1,+\infty )$

$2)\; \; \frac{8-3x}{x+1} \leq \frac{4x+1}{3-x} \\\\ \frac{8-3x}{x+1} -\frac{4x+1}{3-x} \leq 0\\\\ \frac{(8-3x)(3-x)-(4x+1)(x+1)}{(x+1)(3-x)} \leq 0\\\\ \frac{24-17x+3x^2-(4x^2+5x+1)}{(x+1)(3-x)} \leq 0\\\\ \frac{-x^2-22x+23}{(x+1)(3-x)} \leq 0\; ,\; \; \frac{x^2+22x-23}{(x+1)(x-3)} \leq 0\; ,\; \; \frac{(x-1)(x+23)}{(x+1)(x-3)} \leq 0\\\\+++[\, -23\, ]---(-1)+++[\, 1\, ]---(3)+++\\\\x\in [-23,-1)\cup [\, 1,3)$

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228MaDaRa228 Спасибо большое

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