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Question

Cho a + b + c = 0. Chứng minh rằng: $\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)$

ST · Accepted Answer

Ta có: $a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0$$\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0$$\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)$$\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2$ (vì a + b + c = 0)$\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2$  (1)Lại có: $\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)$ (2)Thay (1) vào (2) ta được:$\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\left(đpcm\right)$Câu trả lời: Ta có: $a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0$$\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0$$\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)$$\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2$ (vì a + b + c = 0)$\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2$  (1)Lại có: $\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)$ (2)Thay (1) vào (2) ta được:$\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\left(đpcm\right)$

Dương Lam Hàng · Answer

Ta có: $a+b+c=0\Rightarrow\left(a+b+c\right)^2=0$$\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0$$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)$ (1)Cần chứng minh: $\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)$$\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^4+b^4+c^4\right)$$\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)=a^4+b^4+c^4$$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2$ (Cộng hai vế cho 2(a2b2+b2c2+a2c2)$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left[-2\left(ab+bc+ac\right)\right]^2$ (vì (1))$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+abc^2+a^2bc\right)$$\Leftrightarrow8\left(ab^2c+abc^2+a^2bc\right)=0$<=> 8abc (a+b+c) = 0<=> 0 = 0 (Vì a+b+c = 0 ) (luôn luôn đúng)Vậy => đpcmCâu trả lời: Ta có: $a+b+c=0\Rightarrow\left(a+b+c\right)^2=0$$\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0$$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)$ (1)Cần chứng minh: $\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)$$\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^4+b^4+c^4\right)$$\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)=a^4+b^4+c^4$$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2$ (Cộng hai vế cho 2(a2b2+b2c2+a2c2)$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left[-2\left(ab+bc+ac\right)\right]^2$ (vì (1))$\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+abc^2+a^2bc\right)$$\Leftrightarrow8\left(ab^2c+abc^2+a^2bc\right)=0$<=> 8abc (a+b+c) = 0<=> 0 = 0 (Vì a+b+c = 0 ) (luôn luôn đúng)Vậy => đpcm

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