Hindi, asked by zodenikhil28, 1 month ago

Question No. 7
. Find out selling and
distribution Exp.
Advertisement = Rs. 1000
per month , Salary of
salesmen = 90220 Wages of
shopkeeper = 8880​

Answers

Answered by Anonymous
1

\sf{(x+y)^2={\underline{\underline{\pmb{x^{2}+2xy+y^{2}}}}}}

\sf{(x+y)(x-y)={\underline{\underline{\pmb{x^{2}-y^{2}}}}}}

\large\underline{\sf{Solution-}}

 \sf :\impiles \red{\: {x}^{4}  + \dfrac{1}{ {x}^{4} }  + 1}

 \sf \:  =  \:  \:  \dfrac{ {x}^{8} + 1 +  {x}^{4}  }{ {x}^{4} }

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {x}^{8}  + 2{x}^{4}   + 1 -  {x }^{4} \bigg)

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \pink{ \because  \sf \: On \: adding \: and \: subtracting \:  {x}^{4} }

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {( {x}^{4} + 1) }^{2}  -  { ({x}^{2}) }^{2} \bigg)

  \:  \:  \:  \:  \:

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {x}^{4} + 1 -  {x}^{2}  \bigg) \bigg( {x}^{4} + 1  +   {x}^{2}   \bigg)

  \:  \:  \:  \:  \:  \:  \:  \:  \:

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {x}^{4} + 1 -  {x}^{2}  \bigg) \bigg( {x}^{4}  + 1 +  {2x}^{2} -  {x}^{2}   \bigg)

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \pink{\because \sf \: On \: adding \: and \: subtracting \:  {x}^{2} }\\\\

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {x}^{4} + 1 -  {x}^{2}  \bigg)\bigg(  {( {x}^{2} + 1) }^{2} -   {(x)}^{2}  \bigg)  \\\\

 \sf \:  =  \:  \: \dfrac{1}{ {x}^{4} } \bigg(  {x}^{4} + 1 -  {x}^{2}  \bigg) \bigg(( {x}^{2} + 1 + x)( {x}^{2} + 1 - x)   \bigg)\\\\

 \sf \:  =  \:  \: \bigg( \dfrac{ {x}^{4} + 1 -  {x}^{2}  }{ {x}^{2} } \bigg) \bigg( \dfrac{ {x}^{2} + x + 1 }{x} \bigg) \bigg( \dfrac{ {x}^{2}  - x + 1}{x} \bigg)\\\\

 \sf\red{ \:  =  \:\bigg( {x}^{2}  + \dfrac{1}{ {x}^{2} } - 1  \bigg) \bigg( x + \dfrac{1}{x} + 1 \bigg) \bigg(x +  \dfrac{1}{x} - 1 \bigg)} \\\\

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