Sửa LiOH thành KOH nhé

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Ta có: \(a+b+c=2p\)

\(\Rightarrow b+c=2p-a\)

\(\Leftrightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)

\(\Leftrightarrow b^2+2bc+c^2=4p^2-4ap+a^2\)

\(\Leftrightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\) (1)

Ta có: \(a+b+c=2p\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=4p^2\)

Ta có: \(a+b+c=2pa+b+c\)

`<=>a=2pa`

`@`Giả sử: \(2bc+b^2+c^2-a^2=4p\left(p-2\right)\)

`<=>`\(2bc+b^2+c^2-a^2=4p^2-8p\)

`<=>`\(2bc+b^2+c^2-a^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)-8p\)

\(\Leftrightarrow2a^2+2ab+2ac-8p=0\)

\(\Leftrightarrow2a\left(a+b+c\right)-8p=0\)

\(\Leftrightarrow2a.2p-8p=0\)

\(\Leftrightarrow4ap-8p=0\)

\(\Leftrightarrow4p\left(a-2\right)=0\)

\(\Leftrightarrow a-2=0\)

\(\Leftrightarrow a=2\) (2)

\(\left(1\right);\left(2\right)\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-2\right)\)