\(a^2-b^2-c^2-2bc-2ac-2ab\)

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

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

\(=10^2=100\)

Ủng hộ mik nha thanks nhiều

\(a^2-b^2-c^2-2bc-2ac-2ab\)

\(=a^2-b^2-c^2-2\left(bc+ac+bc\right)\)

\(=\left(a-b-c\right)^2=10^2=100\)

Sửa đề tí nha

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)

-Ta có hằng đẳng thức: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(P=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}+2bc+2ca+2ab\)

\(=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}=\dfrac{2\left(b^3c^3+c^3a^3+a^3b^3\right)}{a^2b^2c^2}=\dfrac{2.\left(ab+bc+ca\right)\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=\dfrac{2.0.\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=0\)

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\(P=a^2-b^2-c^2-2bc-20a\)

\(=a^2-\left(b+c\right)^2-20a\)

\(=a^2-\left(a-10\right)^2-20a\)(vì a + b + c = 10)

\(=a^2-a^2+20a-100-20a\)

\(=-100\)

Ta có kết quả tổng quát hơn như sau:

Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$

Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$