Question

if 3x+1/3x +2= 0, then 243x^7 + 1/81x^2 = ??​

Answer 1

Question

if 3x+1/3x +2= 0, then 243x^7 + 1/81x^2 = ?

Answer

Given:-

${ \boxed{ \rm{ \color{brown}{ \frac{3x + 1}{3x + 2} = 0}}}}$

To find:-

${ \boxed{ \rm{ \color{brown}243x {}^{7} + { \frac{1}{81x {}^{2} } = \_ \_\_ }}}}$

Let's calculate

${ \rm{ \color{maroon}{ \frac{3x + 1}{3x + 2} = 0}}}$

${ \rm{ \color{maroon} \mapsto{ \frac{3x + 1}{3x + 2} = 0}}}$

${ \rm{ \color{maroon} \mapsto{3x + 1}= 0}}$

${ \rm{ \color{maroon} \mapsto{3x }= - 1}}$

${ \rm{ \color{maroon} \mapsto{x}= \frac{ - 1}{ \: \: 3}}}$

Substituting the value of x

${ \rm{ \color{maroon}243x {}^{7} + { \frac{1}{81x {}^{2} } }}}$

${ \rm{ \color{maroon} \mapsto243x {}^{7} + { \frac{1}{81x {}^{2} } }}}$

${ \rm{ \color{maroon} \mapsto243( \frac{ - 1}{3}) {}^{7} + { \frac{1}{81 ( \frac{ - 1}{ \: 3}){}^{2} } }}}$

${ \rm{ \color{maroon} \mapsto243( \frac{ - 1}{2187}) + { \frac{1}{81 ( \frac{ 1}{ \: 9})} }}}$

${ \rm{ \color{maroon} \mapsto \cancel{243}( \frac{ - 1}{ \cancel{2187}}) + { \frac{1}{ \cancel{81} ( \frac{ 1}{ \: \cancel{9}})} }}}$

${ \rm{ \color{maroon} \mapsto { \cancel{81}}( \frac{ - 1}{ \cancel{726}}) \frac{1}{1} \times { { 9}} }}$

${ \rm{ \color{maroon} \mapsto { 27}( \frac{ - 1}{143}) + \frac{1}{1} \times { { 9}} }}$

${ \rm{ \color{maroon} \mapsto( \frac{ - 27}{143}) + { 9}} }$

${ \rm{ \color{maroon} \mapsto( \frac{ - 27 + 1287}{143}) } }$

${ \rm{ \color{maroon} \mapsto\frac{ 1260}{143} } }$

${ \boxed{ \rm{ \color{maroon} \mapsto8.811 } }}$