Предмет: Алгебра,
автор: tutucys
Ответы
Автор ответа:
1
Метод рационализации.
![log_{4-x}(16-x^2) \leq 1\; ,\; \; \; ODZ:\; \left \{ {{4-x\ \textgreater \ 0,\; 4-x\ne 1} \atop {16-x^2\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 4,\; x\ne 3} \atop {x\in (-4,4)}} \right. \; \to \\\\x\in (-4,3)U(3,4)\\\\(4-x-1)(16-x^2-(4-x)) \leq 0\\\\(3-x)(-x^2+x+12) \leq 0\\\\(x-3)(x-4)(x+3) \leq 0\\\\---(-3)+++(3)---(4)+++\\\\x\in (-\infty,-3]U[3,4]\\\\ \left \{ {{x\in (-\infty,-3]U[3,4]} \atop {x\in (-4,3)U(3,4)}} \right. \; \; \to \; \; x\in (-4,3)U(3,4)](https://tex.z-dn.net/?f=log_%7B4-x%7D%2816-x%5E2%29+%5Cleq+1%5C%3B+%2C%5C%3B+%5C%3B+%5C%3B+ODZ%3A%5C%3B++%5Cleft+%5C%7B+%7B%7B4-x%5C+%5Ctextgreater+%5C+0%2C%5C%3B+4-x%5Cne+1%7D+%5Catop+%7B16-x%5E2%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+4%2C%5C%3B+x%5Cne+3%7D+%5Catop+%7Bx%5Cin+%28-4%2C4%29%7D%7D+%5Cright.+%5C%3B+%5Cto+%5C%5C%5C%5Cx%5Cin+%28-4%2C3%29U%283%2C4%29%5C%5C%5C%5C%284-x-1%29%2816-x%5E2-%284-x%29%29+%5Cleq+0%5C%5C%5C%5C%283-x%29%28-x%5E2%2Bx%2B12%29+%5Cleq+0%5C%5C%5C%5C%28x-3%29%28x-4%29%28x%2B3%29+%5Cleq+0%5C%5C%5C%5C---%28-3%29%2B%2B%2B%283%29---%284%29%2B%2B%2B%5C%5C%5C%5Cx%5Cin+%28-%5Cinfty%2C-3%5DU%5B3%2C4%5D%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bx%5Cin+%28-%5Cinfty%2C-3%5DU%5B3%2C4%5D%7D+%5Catop+%7Bx%5Cin+%28-4%2C3%29U%283%2C4%29%7D%7D+%5Cright.+%5C%3B+%5C%3B+%5Cto+%5C%3B+%5C%3B+x%5Cin+%28-4%2C3%29U%283%2C4%29 "log_{4-x}(16-x^2) \leq 1\; ,\; \; \; ODZ:\; \left \{ {{4-x\ \textgreater \ 0,\; 4-x\ne 1} \atop {16-x^2\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 4,\; x\ne 3} \atop {x\in (-4,4)}} \right. \; \to \\\\x\in (-4,3)U(3,4)\\\\(4-x-1)(16-x^2-(4-x)) \leq 0\\\\(3-x)(-x^2+x+12) \leq 0\\\\(x-3)(x-4)(x+3) \leq 0\\\\---(-3)+++(3)---(4)+++\\\\x\in (-\infty,-3]U[3,4]\\\\ \left \{ {{x\in (-\infty,-3]U[3,4]} \atop {x\in (-4,3)U(3,4)}} \right. \; \; \to \; \; x\in (-4,3)U(3,4)")
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