Example

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

Question

$$(\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)})\cdot\frac{a^{2}+2a}{8}:$$

Answer

$$a^2*(2*a^2+4*a+2)+16*a$$

Solution


Simplify  \(1\times a(a+1)\)  to  \(a(a+1)\).
\[(a(a+1)+1\times (a+1)(a+2)){a}^{2}+2a\times 8\]
Simplify  \(1\times (a+1)(a+2)\)  to  \((a+1)(a+2)\).
\[(a(a+1)+(a+1)(a+2)){a}^{2}+2a\times 8\]
Expand.
\[({a}^{2}+a+{a}^{2}+2a+a+2){a}^{2}+2a\times 8\]
Collect like terms.
\[(({a}^{2}+{a}^{2})+(a+2a+a)+2){a}^{2}+2a\times 8\]
Simplify  \(({a}^{2}+{a}^{2})+(a+2a+a)+2\)  to  \(2{a}^{2}+4a+2\).
\[(2{a}^{2}+4a+2){a}^{2}+2a\times 8\]
Regroup terms.
\[{a}^{2}(2{a}^{2}+4a+2)+2a\times 8\]
Simplify  \(2a\times 8\)  to  \(16a\).
\[{a}^{2}(2{a}^{2}+4a+2)+16a\]
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