+118696 (x+1)(2−x)<x(3+x)$Iamgoingtomultiplybothsidesbypositivenumbersandseeifthatwillhelp$(2−x)2(3+x)2∗(x+1)(2−x)<(2−x)2(3+x)2x(3+x)(2−x)(3+x)2(x+1)<(2−x)2(3+x)x(2−x)(3+x)2(x+1)−(2−x)2(3+x)x<0(2−x)(3+x)[(3+x)(x+1)−(2−x)x]<0−(x−2)(x+3)[(x+3)(x+1)−x(−x+2)]<0−(x−2)(x+3)[x2+x+3x+3+x2−2x]<0−(x−2)(x+3)[2x2+2x+3]<0
consider
2x2+2x+3=0△=4−24<0$sincethediscriminantis0therearenorootsforthis.$$theaxisofsymmetryfor$y=2x2+2x+3is−b2a=−24=−12
the roots are x=2 AND x=-3
the polynomial if set to y and graphed will finish in the bottom right corner because of the - out to front.
I can see that the polynomial will be less than 0 when x>2 and when x<-3
I am sorry i probably have not explained this very well, it was a difficult question for this type.
I did not use the graph but I will draw it now to show you.
https://www.desmos.com/calculator/5xfsalfbbg
Feel free to ask questions 
CPhill's way was probably better for this one. :)
+130458
(x+1)/(2-x) < x/(3+x)
This might be solved easiest by some analysis
Let's divide the inequality into three iintervals
(-∞, -3), (-3, 2) and (2, ∞)
Notice that, on the first interval, the function on the left is always less than the function on the right
We can't say what happens at x = -3 because the function on the right isn't defined there
On the second interval, the function on the left is always greater than the function on the right
And at x = 2 the function on the left is undefined
On the third interval, the function on the left is again always less than the function on the right
So, the solution to this problem is that the intervals (-∞, -3) and (2, ∞) make the inequality true and the interval (-3, 2) makes it false
Here's a graph that confirms our suspicions.... https://www.desmos.com/calculator/ymvoluxlsa
+118696 (x+1)(2−x)<x(3+x)$Iamgoingtomultiplybothsidesbypositivenumbersandseeifthatwillhelp$(2−x)2(3+x)2∗(x+1)(2−x)<(2−x)2(3+x)2x(3+x)(2−x)(3+x)2(x+1)<(2−x)2(3+x)x(2−x)(3+x)2(x+1)−(2−x)2(3+x)x<0(2−x)(3+x)[(3+x)(x+1)−(2−x)x]<0−(x−2)(x+3)[(x+3)(x+1)−x(−x+2)]<0−(x−2)(x+3)[x2+x+3x+3+x2−2x]<0−(x−2)(x+3)[2x2+2x+3]<0
consider
2x2+2x+3=0△=4−24<0$sincethediscriminantis0therearenorootsforthis.$$theaxisofsymmetryfor$y=2x2+2x+3is−b2a=−24=−12
the roots are x=2 AND x=-3
the polynomial if set to y and graphed will finish in the bottom right corner because of the - out to front.
I can see that the polynomial will be less than 0 when x>2 and when x<-3
I am sorry i probably have not explained this very well, it was a difficult question for this type.
I did not use the graph but I will draw it now to show you.
https://www.desmos.com/calculator/5xfsalfbbg
Feel free to ask questions
CPhill's way was probably better for this one. :)
