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(х+2)(х+12)·(х+3)(х+8) = 4x²
(x²+14x+24)(x²+11x+24)=4x² | : x²≠0
![frac{x^2+14x+24}{x} * frac{x^2+11x+24}{x} =4\
(x+14+ frac{24}{x} )(x+11+ frac{24}{x} )=4\ \
3AMEHA: x+11+ frac{24}{x} = t\ \
t(t+3)=4\ t^2+3t-4=0\ t_1=1, t_2=-4](https://tex.z-dn.net/?f=+frac%7Bx%5E2%2B14x%2B24%7D%7Bx%7D+%2A+frac%7Bx%5E2%2B11x%2B24%7D%7Bx%7D+%3D4%5C%0A%28x%2B14%2B+frac%7B24%7D%7Bx%7D+%29%28x%2B11%2B+frac%7B24%7D%7Bx%7D+%29%3D4%5C+%5C%0A3AMEHA%3A+x%2B11%2B+frac%7B24%7D%7Bx%7D+%3D+t%5C+%5C%0At%28t%2B3%29%3D4%5C+t%5E2%2B3t-4%3D0%5C+t_1%3D1%2C+t_2%3D-4 "frac{x^2+14x+24}{x} * frac{x^2+11x+24}{x} =4\
(x+14+ frac{24}{x} )(x+11+ frac{24}{x} )=4\ \
3AMEHA: x+11+ frac{24}{x} = t\ \
t(t+3)=4\ t^2+3t-4=0\ t_1=1, t_2=-4")
Вернемся к х:
![1) x+11+ frac{24}{x} =1\
x^2+10x+24=0\
x_1=4; x_2=6](https://tex.z-dn.net/?f=1%29+x%2B11%2B+frac%7B24%7D%7Bx%7D+%3D1%5C%0Ax%5E2%2B10x%2B24%3D0%5C%0Ax_1%3D4%3B+x_2%3D6 "1) x+11+ frac{24}{x} =1\
x^2+10x+24=0\
x_1=4; x_2=6")
![2) x+11+ frac{24}{x} =-4\
x^2+15x+24=0\
x_{3,4}= frac{-15 б sqrt{129} }{2}](https://tex.z-dn.net/?f=2%29+x%2B11%2B+frac%7B24%7D%7Bx%7D+%3D-4%5C%0Ax%5E2%2B15x%2B24%3D0%5C%0Ax_%7B3%2C4%7D%3D+frac%7B-15+%D0%B1++sqrt%7B129%7D+%7D%7B2%7D+ "2) x+11+ frac{24}{x} =-4\
x^2+15x+24=0\
x_{3,4}= frac{-15 б sqrt{129} }{2}")
Ответ:
(x²+14x+24)(x²+11x+24)=4x² | : x²≠0
Вернемся к х:
Ответ:
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