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Cho a+b+c=0 và a2+b2+c2=10. Tính gtbtA = a2(1-a2) + b2(1-b2) + c2(1-c2)

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$a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=0$$\Leftrightarrow10+2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=-5$$\Rightarrow\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=25$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right).0=25$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=25$$a^2+b^2+c^2=10\Leftrightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100$$\Leftrightarrow a^4+b^4+c^4+2.25=100\Leftrightarrow a^4+b^4+c^4=50$$A=a^2\left(1-a^2\right)+b^2\left(1-b^2\right)+c^2\left(1-c^2\right)=a^2+b^2+c^2-\left(a^4+b^4+c^4\right)$$A=10-50=-40$Câu trả lời: $a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=0$$\Leftrightarrow10+2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=-5$$\Rightarrow\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=25$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right).0=25$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=25$$a^2+b^2+c^2=10\Leftrightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100$$\Leftrightarrow a^4+b^4+c^4+2.25=100\Leftrightarrow a^4+b^4+c^4=50$$A=a^2\left(1-a^2\right)+b^2\left(1-b^2\right)+c^2\left(1-c^2\right)=a^2+b^2+c^2-\left(a^4+b^4+c^4\right)$$A=10-50=-40$

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