What is ((3m+1)(m-1))/((m+1)^2)>0 ?

The solution to ((3m+1)(m-1))/((m+1)^2)>0 is m 1

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((3m+1)(m-1))/((m+1)^2)>0

\frac{( 3 m + 1 ) ( m - 1 )}{( m + 1 ) ^{2}} > 0

m < - 1 or - 1 < m < - \frac{1}{3} or m > 1

Interval Notation

(- \infty, - 1) \cup (- 1, - \frac{1}{3}) \cup (1, \infty)

Decimal Notation

m < - 1 or - 1 < m < - 0.33333 \dots or m > 1

Solution steps

\frac{( 3 m + 1 ) ( m - 1 )}{( m + 1 ) ^{2}} > 0

Identify the intervals

m < - 1 or - 1 < m < - \frac{1}{3} or m > 1

2 (10 a - 13) + 5 a = 619

20 + 5

1 \cdot 1 0^{- 5}

4 k^{2} + 12 k - 11 = 5

(63)^{2} + (63)^{2} = c^{2}

What is ((3m+1)(m-1))/((m+1)^2)>0 ?
The solution to ((3m+1)(m-1))/((m+1)^2)>0 is m<-1\lor-1<m<-1/3 \lor m>1