Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { 4 a c - 1.10 m } \\ { ( \frac { 1 } { 12 + a + 8 } ) ^ { - 1 } + ( \frac { 1 } { 10 + c + 1 } ) = } \end{array} \right.$$

Answer

x=(e*t*h*n*(1/(1+a+1/b)+1/(1+b+1/c)+1/(1+c+1/a)))/(b*c)

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[xbc=1\times then(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+{c}^{-1}}+\frac{1}{1+c+{a}^{-1}})\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[xbc=1\times then(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+{a}^{-1}})\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[xbc=1\times then(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})\]
Simplify  \(1\times then(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})\)  to  \(thne(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})\).
\[xbc=thne(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})\]
Regroup terms.
\[xbc=ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})\]
Divide both sides by \(b\).
\[xc=\frac{ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})}{b}\]
Divide both sides by \(c\).
\[x=\frac{\frac{ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})}{b}}{c}\]
Simplify  \(\frac{\frac{ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})}{b}}{c}\)  to  \(\frac{ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})}{bc}\).
\[x=\frac{ethn(\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}})}{bc}\]
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