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How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1? Solution) first thing thats not an inequality,and second thing its very easy if thats the question.the LHS = |10/x^2 + 10/x + 2x + 3| = 110/x^2 will also be an integer only if x^2<10.....i.e. x<3.hence x= 0,+1,-1,+2,-2,+3,-3.but the only value that holds the equality is -1.hence the no. of INTEGERS is only 1.
21-83/4
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what is the lowest term of 11/121
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