Help on questions 11 and 15 Add and simplify if possible

Mathematics

2 answers:

Svetllana [295]2 years ago

7 0

$\frac{16+x}{x^3}+\frac{7-4x}{x^3}=\frac{16+x+7-4x}{x^3}=\frac{23-3x}{x^3}\\\\======================================\\\\\frac{5}{t-1}+\frac{3}{t}=\frac{5t}{t(t-1)}+\frac{3(t-1)}{t(t-1)}=\frac{5t+3t-3}{t^2-t}=\frac{8t-3}{t^2-t}$

nika2105 [10]2 years ago

4 0

$\frac{16+x}{x^{3} } + \frac{7-4x}{ x^{3} } 
 \\ 16+x+7-4x = 23-3x
 \\ \frac{23 -3x}{ x^{3} }$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\frac{5}{t-1} +\frac{3}{t} 
 \\ \\ \frac{5t+(t-1)(3)}{(t-1)(t)} 
 \\ \\ \frac{5t+3t-3}{ t^{2}-t } 
 \\ \\ \frac{8t-3}{ t^{2}-t }$

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How can we factorise the expression: 2x^6+17x^3+8?

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$(a + 2)( {a}^{2} - 2a + 4 )(2 {x}^{ 3} + 1)$

Step-by-step explanation:

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