Solve 6atleft(1+t^2right)-3at^2.2t/left(1+t^2right)^2/3aleft(1+{t^2right)-3at.2t/{left(1+{t^2right)}^2}}

$$\frac{ \frac{ 6at \left( 1+ { t }^{ 2 } \right) -3a { t }^{ 2 } .2t }{ { \left(1+ { t }^{ 2 } \right) }^{ 2 } } }{ \frac{ 3a \left( 1+ { t }^{ 2 } \right) -3at.2t }{ { \left(1+ { t }^{ 2 } \right) }^{ 2 } } }$$

Evaluate

$$\frac{t\left(9t^{2}+10\right)}{4t^{2}+5}$$

Short Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{\frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}}}{\frac{3a\left(1+t^{2}\right)-3at\times 0.2t}{\left(1+t^{2}\right)^{2}}}$$

Multiply t and t to get t^{2}.

$$\frac{\frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}}}{\frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}}}$$

Divide \frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}} by \frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}} by multiplying \frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}} by the reciprocal of \frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}}.

$$\frac{\left(6at\left(1+t^{2}\right)-3at^{3}\times 0.2\right)\left(1+t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}\left(3a\left(1+t^{2}\right)-3at^{2}\times 0.2\right)}$$

Cancel out \left(t^{2}+1\right)^{2} in both numerator and denominator.

$$\frac{6at\left(t^{2}+1\right)-0.2\times 3at^{3}}{3a\left(t^{2}+1\right)-0.2\times 3at^{2}}$$

Multiply 0.2 and 3 to get 0.6.

$$\frac{6at\left(t^{2}+1\right)-0.6at^{3}}{3a\left(t^{2}+1\right)-0.2\times 3at^{2}}$$

Multiply 0.2 and 3 to get 0.6.

$$\frac{6at\left(t^{2}+1\right)-0.6at^{3}}{3a\left(t^{2}+1\right)-0.6at^{2}}$$

Factor the expressions that are not already factored.

$$\frac{0.6at\left(9t^{2}+10\right)}{0.6a\left(4t^{2}+5\right)}$$

Cancel out a in both numerator and denominator.

$$\frac{0.6t\left(9t^{2}+10\right)}{0.6\left(4t^{2}+5\right)}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{t\left(9t^{2}+10\right)}{0.6^{0}\left(4t^{2}+5\right)}$$

Expand the expression.

$$\frac{9t^{3}+10t}{4t^{2}+5}$$

Expand

$$\frac{9t^{3}+10t}{4t^{2}+5}$$

Short Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{\frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}}}{\frac{3a\left(1+t^{2}\right)-3at\times 0.2t}{\left(1+t^{2}\right)^{2}}}$$

Multiply t and t to get t^{2}.

$$\frac{\frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}}}{\frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}}}$$

Divide \frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}} by \frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}} by multiplying \frac{6at\left(1+t^{2}\right)-3at^{3}\times 0.2}{\left(1+t^{2}\right)^{2}} by the reciprocal of \frac{3a\left(1+t^{2}\right)-3at^{2}\times 0.2}{\left(1+t^{2}\right)^{2}}.

$$\frac{\left(6at\left(1+t^{2}\right)-3at^{3}\times 0.2\right)\left(1+t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}\left(3a\left(1+t^{2}\right)-3at^{2}\times 0.2\right)}$$

Cancel out \left(t^{2}+1\right)^{2} in both numerator and denominator.

$$\frac{6at\left(t^{2}+1\right)-0.2\times 3at^{3}}{3a\left(t^{2}+1\right)-0.2\times 3at^{2}}$$

Multiply 0.2 and 3 to get 0.6.

$$\frac{6at\left(t^{2}+1\right)-0.6at^{3}}{3a\left(t^{2}+1\right)-0.2\times 3at^{2}}$$

Multiply 0.2 and 3 to get 0.6.

$$\frac{6at\left(t^{2}+1\right)-0.6at^{3}}{3a\left(t^{2}+1\right)-0.6at^{2}}$$

Factor the expressions that are not already factored.

$$\frac{0.6at\left(9t^{2}+10\right)}{0.6a\left(4t^{2}+5\right)}$$

Cancel out a in both numerator and denominator.

$$\frac{0.6t\left(9t^{2}+10\right)}{0.6\left(4t^{2}+5\right)}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{t\left(9t^{2}+10\right)}{0.6^{0}\left(4t^{2}+5\right)}$$

Expand the expression.

$$\frac{9t^{3}+10t}{4t^{2}+5}$$