Question

Evaluate x^p/x^p + x^q + 1/xp-q + 1​

Answer 1

Evaluate

• $\sf\dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{x^{p-q }+ 1}$

Solution

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{x^{p-q }+ 1}}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{(x^{p} \times {x}^{ - q} +) + 1}}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{ \bigg(x^{p} \times \dfrac{1}{ {x}^{ - q} } \bigg) + 1}}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{ \dfrac{ {x}^{p} }{ {x}^{q} } + 1}}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{ \dfrac{ {x}^{p} }{ {x}^{q} } + 1}}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{ \dfrac{ {x}^{p} + {x}^{q} }{ {x}^{q} } }}$

${ \sf~~~~~\implies \dfrac{x^p}{x^p + x^q} +\dfrac{ {x}^{q} }{ {x}^{p} + {x}^{q} }}$

${ \sf~~~~~\implies { \cancel{\dfrac{x^p + {x}^{q} }{x^p + x^q} }}}^{1}$

${ \sf~~~~~\implies 1}$

$\huge~~~~~~~~~~~~~~~~~~~~~\\\hline{ }$

$\sf\dfrac{x^p}{x^p + x^q} +\dfrac{ 1}{x^{p-q }+ 1} \: \bf= 1$

$\huge~~~~~~~~~~~~~~~~~~~~~\\\hline{ }$

• $\gray{\tiny\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{}}\\ {\boxed{\begin{array}{c | c} \frac{ \: ~~~~~~~~~~\: \: \: \: \:\sf Laws \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{ } &\frac{ \: ~~~~~~~~~~ \: \: \: \: \:\sf Example \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{ }\\ \sf \bigstar{a}^{m} \times {a}^{n} = {a}^{m + n} & \sf {a}^{2} \times {a}^{3} = {a}^{2 + 3} = {a}^{6} \\ \\ \sf \bigstar{a}^{m} \div {a}^{n} = {a}^{m - n}& \sf {a}^{3} \div {a}^{2} = {a}^{3 - 2} = {a}^{1} \\ \\ \sf{\bigstar \: \: \: \: \: \: ( {a}^{m} ) ^{n} = {a}^{mn} } & \sf( {a}^{2} ) ^{3} = {a}^{2 \times 3} = {a}^{6} \\ \\ {\bigstar\sf a {}^{m} \times {n}^{m} = (ab) ^{m} } &\sf a {}^{2} \times {b}^{2} = (ab) ^{2}\\ \\ \sf\bigstar \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {a}^{0} = 1& \sf {2}^{0} = 1 \: \: \: \: \\ \\ \sf \bigstar \: \: \: \: {\dfrac{ {a}^{m} }{ {b}^{m} }= \left( \dfrac{a}{b} \right) ^{m} }& \sf{\dfrac{ {a}^{2} }{ {b}^{2} }= \left( \dfrac{a}{b} \right) ^{2} }\\\\\bigstar~~~~~~~ \sf x^{\frac{m}{n} }=\sqrt[n]{x^m}\sf = (\sqrt[n]{x})^m & \sf x^{\frac{2}{3} }=\sqrt[3]{x^2} = (\sqrt[n]{x})^m\\ \\\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}$

Answer 2

$\huge\boxed{\fcolorbox{blue}{lime}{question✨:}}$

$\tt\dfrac\red{x^p}\red{x^p + x^q} +\dfrac\red{ 1}\red{x^{p-q }+ 1}$

$\huge\pink{\boxed{\green {\mathbb{\overbrace {\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{†◖ \: ⚆SOLTUON⚆◗†}}}}}}}}}$

${ \tt~~~~~\implies \dfrac\blue{x^p}\blue{x^p + x^q} +\dfrac\blue{ 1}\blue{x^{p-q }+ 1}}$

${\tt~~~~~\implies \dfrac \pink{x^p} \pink{x^p + x^q} +\dfrac \pink{ 1} \pink{(x^{p} \times {x}^{ - q} +) + 1}}$

${ \tt~~~~~\implies \dfrac \orange{x^p} \orange{x^p + x^q} +\dfrac \orange{ 1} \orange{ \bigg(x^{p} \times \dfrac \orange{1} \orange{ \orange \orange{x}^ \orange{ - q} } \bigg) \orange + \orange1}}$

${ \sf~~~~~\implies \dfrac\red{x^p}\red{x^p + x^q} +\dfrac\red{ 1}\red{ \dfrac \red{ {x}^{p} }\red{ {x}^{q} } + 1}}$

${ \tt~~~~~\implies \dfrac\purple{x^p} \purple{x^p + x^q} +\dfrac \purple{ 1}{ \dfrac \purple{ \purple{x}^{p} } \purple{ {x}^{q} } \purple+ \purple1}}$

${ \mathtt~~~~~\implies \dfrac \green{x^p} \green{x^p + x^q} +\dfrac \green{ 1}{ \dfrac \green{ {x}^{p} + {x}^{q} } \green{ {x}^{q} } }}$

${ \textsf~~~~~\implies \dfrac \red{x^p} \pink{x^p + x^q} +\dfrac \blue{ {x}^{q} } \purple{ {x}^{p} + {x}^{q} }}$

${ \sf~~~~~\implies { \cancel{\dfrac \pink{x^p + {x}^{q} } \blue{x^p + x^q} }}}^{1}$

$\begin{gathered} \blue{ \tt~~~~~\implies 1} \\ \\ \\ \end{gathered}$

$\tt \huge\dfrac \purple{x^p} \pink{x^p + x^q} +\dfrac \blue{ 1} \green{x^{p-q }+ 1} \: \bf$

$\huge\pink{\boxed{\green{\mathtt{\overbrace{\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{⚘thanks ⚘}}}}}}}}}$

${\huge{\boxed{\tt{\color {red}{hope \: it \: helps \: you❀✿°᭄}}}}}$

${\Huge{ \red{\underline{\underline{\bf{\maltese \pink{ \: @Rosoni28 }}}}}}}$

$\huge \color{blue} \boxed{\colorbox{lime}{------------×-----------}}$