Câu hỏi của Olala Thành

Question

Cho a + b + c = 0. Chứng minh a^4 + b^4 + c^4 bằng mỗi biểu thức:a) 2(a^2b^2 + b^2c^2 + c^2a^2)b) 2( ab + bc + ca)^2c) (a^2 + b^2 + c^2)^2 / 2

Dark Killer · Accepted Answer

a) Ta có: $a+b+c=0$$\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0$$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)$$\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)$$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]$$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)$$\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)$$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)$Câu trả lời: a) Ta có: $a+b+c=0$$\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0$$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)$$\Rightarrow 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+2a^2bc+2ab^2c+2abc^2\right)$$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]$$\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)$$\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)$$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)$

Dark Killer · Answer

b) Ta có: $a+b+c=0$$\Rightarrow2abc\left(a+b+c\right)=0$$\Rightarrow2a^2bc+2ab^2c+2abc^2=0$Ta lại có:$a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2$(chứng minh câu a)$\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2$$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)$$\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2$Câu trả lời: b) Ta có: $a+b+c=0$$\Rightarrow2abc\left(a+b+c\right)=0$$\Rightarrow2a^2bc+2ab^2c+2abc^2=0$Ta lại có:$a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2$(chứng minh câu a)$\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2$$\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)$$\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2$

Dark Killer · Answer

c) Ta có: $a+b+c=0$$\Rightarrow a=-\left(b+c\right)$$\Rightarrow a^2=b^2+c^2+2bc$$\Rightarrow a^2-b^2-c^2=2bc$$\Rightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2=4b^2c^2$$\Rightarrow a^4+b^4+c^4=4b^2c^2+2a^2b^2+2a^2c^2-2b^2c^2$$\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2$$\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$$\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2$$\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right):2$(Nhớ k cho mình với nhá!)Câu trả lời: c) Ta có: $a+b+c=0$$\Rightarrow a=-\left(b+c\right)$$\Rightarrow a^2=b^2+c^2+2bc$$\Rightarrow a^2-b^2-c^2=2bc$$\Rightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2=4b^2c^2$$\Rightarrow a^4+b^4+c^4=4b^2c^2+2a^2b^2+2a^2c^2-2b^2c^2$$\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2$$\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$$\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2$$\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right):2$(Nhớ k cho mình với nhá!)

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