logx/3(logx(√3-x))≥0
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Ответы на вопрос
Ответил IrkaShevko
0
ОДЗ: 0 < x < 3; x ≠ 1
=> x/3 < 1
![log_x sqrt{3-x} leq 1\
1) 0 textless x textless 1\
sqrt{3 - x} geq x\
3 - x geq x^2\
x^2 + x - 3 leq 0\
D=1 + 12 = 13\
x_1 = frac{ sqrt{13} -1}{2} textgreater frac{3-1}{2} =1\
x_2 = frac{- sqrt{13}-1 }{2} textless 0\
0 textless x textless 1\\
2) 1 textless x textless 3\
sqrt{3 - x} leq x\
3 - x leq x^2\
x^2 + x - 3 geq 0\
x_1 = frac{ sqrt{13} -1}{2} textgreater frac{3-1}{2} =1\
x_1 textless 3\
x_2 = frac{- sqrt{13}-1 }{2} textless 0\
frac{ sqrt{13} -1}{2} leq x textless 3](https://tex.z-dn.net/?f=log_x+sqrt%7B3-x%7D++leq++1%5C%0A1%29+0++textless+++x++textless+++1%5C%0A+sqrt%7B3+-+x%7D+geq++x%5C%0A3+-+x++geq++x%5E2%5C%0Ax%5E2+%2B+x+-+3+leq+0%5C%0AD%3D1+%2B+12+%3D+13%5C%0Ax_1+%3D++frac%7B+sqrt%7B13%7D+-1%7D%7B2%7D+textgreater+++frac%7B3-1%7D%7B2%7D++%3D1%5C%0Ax_2+%3D++frac%7B-+sqrt%7B13%7D-1+%7D%7B2%7D++textless++0%5C%0A0++textless+++x++textless+++1%5C%5C%0A2%29+1++textless+++x++textless+++3%5C%0A+sqrt%7B3+-+x%7D+leq+x%5C%0A3+-+x+leq+x%5E2%5C%0Ax%5E2+%2B+x+-+3+geq+0%5C%0Ax_1+%3D++frac%7B+sqrt%7B13%7D+-1%7D%7B2%7D+textgreater+++frac%7B3-1%7D%7B2%7D++%3D1%5C%0Ax_1++textless+++3%5C%0Ax_2+%3D++frac%7B-+sqrt%7B13%7D-1+%7D%7B2%7D++textless++0%5C%0Afrac%7B+sqrt%7B13%7D+-1%7D%7B2%7D++leq+x++textless++3 "log_x sqrt{3-x} leq 1\
1) 0 textless x textless 1\
sqrt{3 - x} geq x\
3 - x geq x^2\
x^2 + x - 3 leq 0\
D=1 + 12 = 13\
x_1 = frac{ sqrt{13} -1}{2} textgreater frac{3-1}{2} =1\
x_2 = frac{- sqrt{13}-1 }{2} textless 0\
0 textless x textless 1\\
2) 1 textless x textless 3\
sqrt{3 - x} leq x\
3 - x leq x^2\
x^2 + x - 3 geq 0\
x_1 = frac{ sqrt{13} -1}{2} textgreater frac{3-1}{2} =1\
x_1 textless 3\
x_2 = frac{- sqrt{13}-1 }{2} textless 0\
frac{ sqrt{13} -1}{2} leq x textless 3")
Ответ: x ∈ (0; 1) U [(√13 - 1)/2; 3)
=> x/3 < 1
Ответ: x ∈ (0; 1) U [(√13 - 1)/2; 3)
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