7-(2/b)(5/b)

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Solution for 7-(2/b)(5/b) equation:


D( b )

b = 0

b = 0

b = 0

b in (-oo:0) U (0:+oo)

7-(2/b) < 5/b // - 5/b

7-(2/b)-(5/b) < 0

7-2*b^-1-5*b^-1 < 0

-7*b^-1 < -7

-7*b^-1 < -7 // * -1

7*b^-1 > 7

7*b^-1 > 7 // : 7

b^-1 > 7/7

b^-1 > 1

1/(b^1) > 1

b <> 0

1/(b^1) > 1 // * b^2

(b^2)/(b^1) > 1*b^2

b^1 > 1*b^2

b-b^2 > 0

b^1*(1-b) > 0

b < 0

-(abs(b))^1*(1*(abs(b))^1+1) > 0 // : -(abs(b))^1

1*(abs(b))^1+1 < 0

1*(abs(b))^1 < -1 // : 1

(abs(b))^1 < -1/1

(abs(b))^1 < -1 // ^ 1/1

abs(b) < (-1)^(1/1)

abs(b) < -1

b belongs to the empty set

b < 0

b > 0

b^1*(1-(1*b^1)) > 0 // : b^1

1-(1*b^1) > 0

1 > 1*b^1 // : 1

1/1 > b^1 // ^ 1/1

(1/1)^(1/1) > b

b in (0:1)

b > 0

b in (0:1)

b in (0:1)

(0:1)

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