3 years ago

3(x+3/4+5-x/4-x/6)=10

Mathematics

1 answer:

viva [34]3 years ago

5 0

3(x+3-4+5-x-4-x-6)=10

-3x/-3= -28/3

x= -28/3

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Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)

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GIVEN:

$5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)$

remember:

$\sqrt{u} = {u}^{ \frac{1}{2} }$

And

${u}^{n} \times {u}^{m} = {u}^{n + m}$

SOLVE:

start by multiplying the factors:

$5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))$

simplify by combing like terms. Most terms subtract off, leaving:

$5( {u}^{6} - 1)$

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

$5( {u}^{3} - 1)( {u}^{3} + 1)$

I'm not exactly sure what form they want the answer in...

so taking the square root:

$\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }$

so my best answer is:

${5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }$
or the more factored form:

${5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }$

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!

7 0

3 years ago

3(x+3/4+5-x/4-x/6)=10​

3(x+3/4+5-x/4-x/6)=10