Math, asked by krushnakajale2007, 20 days ago

-9/12 - X = 10/12 find the value of x​

Answers

Answered by Anonymous
50

\huge\boxed{\fcolorbox{black}{pink} {Solution :-}}

 =  >  \frac{ - 9}{12}  - x =  \frac{10}{12}  \\  =  >  - x =  \frac{10}{12}  +  \frac{9}{12}  \\  =  >  - x =  \frac{19}{12}  \\  =  > x =  \frac{ - 19}{12}  =-1  \frac{7}{12}

Answered by LaRouge
5

Answer:

Answer :-

The value of \sf x^{3} +\sf\cfrac{1}{{x}^{3}}x

3

+

x

3

1

= 488

To Find :-

The value of \sf x^{3} +\cfrac{1}{{x}^{3}}x

3

+

x

3

1

Given :-

\sf x + \cfrac{1}{x} = 8x+

x

1

=8

Step By Step Explanation :-

We know that \sf{x + \cfrac{1}{x} = 8}x+

x

1

=8

We need to calculate the value of \sf x^{3} +\cfrac{1}{{x}^{3}}x

3

+

x

3

1

So let's do it !!

\begin{gathered} \bf \: Using \: Identity \downarrow \\ \\\dag\boxed{ \bf{ \red {{(x + y)}^{3}= {x}^{3} + {y}^{3} + 3xy(x + y)}}}\end{gathered}

UsingIdentity↓

(x+y)

3

=x

3

+y

3

+3xy(x+y)

By substituting the values ⤵

\begin{gathered} \sf \left(\cfrac{1}{x}+x\right)^{3}= \cfrac{1}{ {x}^{3} } + {x}^{3} + 3 \times \cfrac{1}{ \not x} \times \not x \: \left( x + \cfrac{1}{x}\right) \\ \\ \bf \: By \: substituting \: the \: values \downarrow \\ \\\implies \sf {(8)}^{3} = \cfrac{1}{ {x}^{3} } + {x}^{3} + 3(8) \\ \\\implies \sf 512 = \cfrac{1}{ {x}^{3} } + {x}^{3} + 24 \\ \\ \implies \sf512 - 24 = \cfrac{1}{ {x}^{3} } + {x}^{3} \\ \\ \implies\bf 488 = \cfrac{1}{ {x}^{3} } + {x}^{3} \end{gathered}

(

x

1

+x)

3

=

x

3

1

+x

3

+3×

x

1

×

x(x+

x

1

)

Bysubstitutingthevalues↓

⟹(8)

3

=

x

3

1

+x

3

+3(8)

⟹512=

x

3

1

+x

3

+24

⟹512−24=

x

3

1

+x

3

⟹488=

x

3

1

+x

3

Hence the value of \sf x^{3} +\cfrac{1}{{x}^{3}}x

3

+

x

3

1

= 488

__________________________

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