What is (1^2+4+1)/((1^2+1+1)^2) ?

The solution to (1^2+4+1)/((1^2+1+1)^2) is 2/3

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(1^2+4+1)/((1^2+1+1)^2)

\frac{1 ^{2} + 4 + 1}{( 1 ^{2} + 1 + 1 ) ^{2}}

\frac{2}{3}

Solution steps

\frac{1 ^{2} + 4 + 1}{( 1 ^{2} + 1 + 1 ) ^{2}}

Solve

1^{2} + 4 + 1 : 6

Solve

(1^{2} + 1 + 1)^{2} : 9

= \frac{6}{9}

Cancel the common factor:

3

= \frac{2}{3}

(x + 6) (x - 2) (x - 5) < 0

58 - 10 x \leq 20 + 9 x

\frac{2}{3} + \frac{1}{4}

15 z^{2} - 28 z + 12

(y + 2)^{2} - 14 (y + 2) = - 24

What is (1^2+4+1)/((1^2+1+1)^2) ?
The solution to (1^2+4+1)/((1^2+1+1)^2) is 2/3