Question

help me to solve this, find the real solution
36x^(-(4)/(3))-13x(-2)/(3)+1=0

Answer 1

@518nad help me

Answer 2

@jim_thompson5910

Answer 3

\[36 x ^{\frac{ -4 }{ 3 }}-13 x ^{\frac{ -2 }{ 3 }}+1=0\]
Am I correct?

Answer 4

If i am correct ,then put \[x ^{\frac{ -2 }{ 3 }}=t\]
squaring
\[\left( x ^{\frac{ -2 }{ 3 }} \right)^2=t^2,x ^{\frac{ -4 }{ 3 }}=t^2\]
\[36 t^2-13t+1=0\]
\[36 t^2-9 t-4 t+1=0\]
\[9 t(4 t-1)-1(4t-1)=0\]
\[(4t-1)(9t-1)=0,t=\frac{ 1 }{ 4 },\frac{ 1 }{ 9 }\]
\[t=\frac{ 1 }{ 4 },x ^{\frac{ -2 }{ 3 }}=\frac{ 1 }{ 4 }\]
\[\frac{ 1 }{ x ^{\frac{ 2 }{ 3 }} }=\frac{ 1 }{ 4 },\]
\[x ^{\frac{ 2 }{ 3 }}=4 \]
cubing both sides
\[\left( x ^{\frac{ 2 }{ 3 }} \right)^3=4^3=64,x^2=64,x=\pm 8\]
similarly when \[t=\frac{ 1 }{ 9 },x ^{\frac{ -2 }{ 3 }}=\frac{ 1 }{ 9 },\frac{ 1 }{ x ^{\frac{ 2 }{ 3 }} }=\frac{ 1 }{ 9 }\]
\[x ^{\frac{ 2 }{ 3 }}=9,cubing\]
\[x^2=9^3=729,x= \pm 27\]

Answer 5

thank you so much

Answer 6

@sshayer help me please

Answer 7

help on this one i know, you know how to solve this one.
1)(1)/(1)+2(x+1)(1)/(5)=2

Answer 8

3(x+1)(1)/(1)+2(x+1)(1)/(5)=2

Answer 9

help me please

Answer 10

your statement is ambiguous
is it \[\frac{ 3\left( x+1 \right) }{ 1+2\left( x+1 \right)\frac{ 1 }{ 5 } }?\]

Answer 11

\[3(x+1)\frac{ 1 }{ 1 }+2(x+1)\frac{ 1 }{ 5 }=2\]

Answer 12

help on this one please?

Answer 13

@brooke..help00

Answer 14

\[\left( x+1 \right)\left( 3+\frac{ 2 }{ 5 } \right)=2\]
\[\left( x+1 \right)\left( \frac{ 15+2 }{ 5 } \right)=2\]
\[x+1=2 \times \frac{ 5 }{ 17 }=\frac{ 10 }{ 17 }\]
\[x=\frac{ 10 }{ 17 }-1=?\]