\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)
Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)
\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)
\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)
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