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ОДЗ: х≠2
х≠3
![\frac{1}{x-2}+ \frac{1}{3-x}-5 \leq 0 \\ \\ \frac{1}{x-2}- \frac{1}{x-3}- \frac{5(x-2)(x-3)}{(x-2)(x-3)} \leq 0 \\ \\ \frac{x-3-(x-2)-5(x^2-2x-3x+6)}{(x-2)(x-3)} \leq 0 \\ \\ \frac{x-3-x+2-5x^2+25x-30}{(x-2)(x-3)} \leq 0 \\ \\ \frac{-5x^2+25x-31}{(x-2)(x-3)} \leq 0 \\ \\ \frac{-(5x^2-25x+31)}{(x-2)(x-3)} \leq 0 \\ \\ \frac{5x^2-25x+31}{(x-2)(x-3)} \geq[tex]5x^2-25x+31=5(x- \frac{25- \sqrt{5} }{10} )(x- \frac{25+ \sqrt{5} }{10} ) \\ \\ (x- \frac{25- \sqrt{5} }{10} )(x- \frac{25+ \sqrt{5} }{10} )(x-2)(x-3) \geq0 \\ \\ ](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7Bx-2%7D%2B+%5Cfrac%7B1%7D%7B3-x%7D-5+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7B1%7D%7Bx-2%7D-+%5Cfrac%7B1%7D%7Bx-3%7D-+%5Cfrac%7B5%28x-2%29%28x-3%29%7D%7B%28x-2%29%28x-3%29%7D+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7Bx-3-%28x-2%29-5%28x%5E2-2x-3x%2B6%29%7D%7B%28x-2%29%28x-3%29%7D+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7Bx-3-x%2B2-5x%5E2%2B25x-30%7D%7B%28x-2%29%28x-3%29%7D+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7B-5x%5E2%2B25x-31%7D%7B%28x-2%29%28x-3%29%7D+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7B-%285x%5E2-25x%2B31%29%7D%7B%28x-2%29%28x-3%29%7D+%5Cleq+0+%5C%5C++%5C%5C+%0A+%5Cfrac%7B5x%5E2-25x%2B31%7D%7B%28x-2%29%28x-3%29%7D+++++++++++%5Cgeq%5Btex%5D5x%5E2-25x%2B31%3D5%28x-+%5Cfrac%7B25-+%5Csqrt%7B5%7D+%7D%7B10%7D+%29%28x-+%5Cfrac%7B25%2B+%5Csqrt%7B5%7D+%7D%7B10%7D+%29+%5C%5C++%5C%5C+%0A%28x-+%5Cfrac%7B25-+%5Csqrt%7B5%7D+%7D%7B10%7D+%29%28x-+%5Cfrac%7B25%2B+%5Csqrt%7B5%7D+%7D%7B10%7D+%29%28x-2%29%28x-3%29+%5Cgeq0+%5C%5C++%5C%5C+%0A++)
0[/tex]
Разложим на множители:
5x²-25x+31=0
D=625-4*5*31=625-620=5
x₁=(25-√5)/10≈2.27
x₂=(25+√5)/10≈2.72
+ - + - +
------- 2 --------- (25-√5)/10 ----------- (25+√5)/10 -------- 3 ----------
\\\\\\\\\ \\\\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; 2)U[(25-√5)/10; (25+√5)/10]U(3; +∞)
х≠3
Разложим на множители:
5x²-25x+31=0
D=625-4*5*31=625-620=5
x₁=(25-√5)/10≈2.27
x₂=(25+√5)/10≈2.72
+ - + - +
------- 2 --------- (25-√5)/10 ----------- (25+√5)/10 -------- 3 ----------
\\\\\\\\\ \\\\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; 2)U[(25-√5)/10; (25+√5)/10]U(3; +∞)