Ta có \(7a^2-15ab+2b^2=0\Leftrightarrow7a^2-14ab-ab+2b^2=0\Leftrightarrow7a\left(a-2b\right)-b\left(a-2b\right)=0\Leftrightarrow\left(a-2b\right)\left(7a-b\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}7a-b=0\\a-2b=0\end{matrix}\right.\)(*)

Vì a-2b\(\ne0\)(Để E xác định)

Vậy (*)\(\Leftrightarrow7a-b=0\Leftrightarrow7a=b\)

Thay vào E ta có \(E=\dfrac{a-7a}{2a+7a}-\dfrac{3a-7a}{a-14a}=\dfrac{-6a}{9a}-\dfrac{-4a}{-13a}=\dfrac{-6}{9}-\dfrac{4}{13}=-\dfrac{38}{39}\)

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\(P=2\left(a^3+b^3\right)+7ab\left(a+b\right)\)

\(=2\left(a+b\right)\left(a^2-ab+b^2\right)+7ab\left(a+b\right)\)

\(=\left(a+b\right)\left(2a^2-2ab+2b^2+7ab\right)\)

\(=\left(a+b\right)\left(2a^2+5ab+2b^2\right)\)

\(=\left(a+b\right)\left[\left(2a^2+ab\right)+\left(4ab+2b^2\right)\right]\)

\(=\left(a+b\right)\left[a\left(2a+b\right)+2b\left(2a+b\right)\right]\)

\(=\left(a+b\right)\left(2a+b\right)\left(2b+a\right)\)

= 2( a^3 + b^3 ) + 7ab(a+b) = 2(a+b)(a^2 -ab +b^2)  + 7ab(a+b)  = (a+b) ( 2a^2 - 2ab + 2b^2 - 7ab)

=(a +b ) ( 2a^2 +2b^2 - 9ab)

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