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Решим с помощью метода интервалов.
![1) (x-2)^2(x+1)(x-1)^2(x^2+2x+5) textless 0 \ \
(x^2+2x+5)=0 \ \
D=4-4*1*5 textless 0 \ \
Znachit ;; ne ;; uchitovaem \ \
+++++[-1]-----[1^p]-----[2]+++++ \ \
boxed {xin (-1;1)U(1;2)} \ \ \
2) frac{(x-3)^2(x-2)x}{(x+1)^4(x+5)} textgreater 0 \ \
---[-5]+++[-1^p]+++[0]---[-2]+++[3^p]+++ \ \
boxed {xin (-5;-1)U(-1;0)U(2;3)U(3;+infty)} \ \ \
3) sqrt{x} (x+2) textgreater 0 \ \
ODZ:;x geq 0 \ \
x textgreater -2-ne ;;vxodit ;; v ;; ODZ, ;; znachi: \ \
boxed {xin (0;+infty)}](https://tex.z-dn.net/?f=1%29+%28x-2%29%5E2%28x%2B1%29%28x-1%29%5E2%28x%5E2%2B2x%2B5%29+textless++0+%5C++%5C+%0A%28x%5E2%2B2x%2B5%29%3D0+%5C++%5C+%0AD%3D4-4%2A1%2A5+textless++0+%5C++%5C+%0AZnachit+%3B%3B+ne+%3B%3B+uchitovaem++%5C++%5C+%0A%2B%2B%2B%2B%2B%5B-1%5D-----%5B1%5Ep%5D-----%5B2%5D%2B%2B%2B%2B%2B+%5C++%5C+%0Aboxed+%7Bxin+%28-1%3B1%29U%281%3B2%29%7D+%5C++%5C++%5C+%0A2%29++frac%7B%28x-3%29%5E2%28x-2%29x%7D%7B%28x%2B1%29%5E4%28x%2B5%29%7D+textgreater++0+%5C++%5C+%0A---%5B-5%5D%2B%2B%2B%5B-1%5Ep%5D%2B%2B%2B%5B0%5D---%5B-2%5D%2B%2B%2B%5B3%5Ep%5D%2B%2B%2B+%5C++%5C+%0Aboxed+%7Bxin+%28-5%3B-1%29U%28-1%3B0%29U%282%3B3%29U%283%3B%2Binfty%29%7D++%5C++%5C++%5C+%0A3%29++sqrt%7Bx%7D+%28x%2B2%29+textgreater++0+%5C++%5C+%0AODZ%3A%3Bx+geq+0+%5C++%5C+%0Ax+textgreater++-2-ne+%3B%3Bvxodit+%3B%3B+v+%3B%3B+ODZ%2C+%3B%3B+znachi%3A+%5C++%5C%0Aboxed+%7Bxin+%280%3B%2Binfty%29%7D "1) (x-2)^2(x+1)(x-1)^2(x^2+2x+5) textless 0 \ \
(x^2+2x+5)=0 \ \
D=4-4*1*5 textless 0 \ \
Znachit ;; ne ;; uchitovaem \ \
+++++[-1]-----[1^p]-----[2]+++++ \ \
boxed {xin (-1;1)U(1;2)} \ \ \
2) frac{(x-3)^2(x-2)x}{(x+1)^4(x+5)} textgreater 0 \ \
---[-5]+++[-1^p]+++[0]---[-2]+++[3^p]+++ \ \
boxed {xin (-5;-1)U(-1;0)U(2;3)U(3;+infty)} \ \ \
3) sqrt{x} (x+2) textgreater 0 \ \
ODZ:;x geq 0 \ \
x textgreater -2-ne ;;vxodit ;; v ;; ODZ, ;; znachi: \ \
boxed {xin (0;+infty)}")
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