Câu hỏi của Arata

Question

CMR:$\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Dũng Nguyễn · Accepted Answer

Ta có:$\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca$

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

$=a^2+2ab+b^2+2ac+2bc+c^2$

$=a^2+b^2+c^2+2ab+2bc+2ca$ (đpcm)Câu trả lời: Ta có:$\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca$

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

$=a^2+2ab+b^2+2ac+2bc+c^2$

$=a^2+b^2+c^2+2ab+2bc+2ca$ (đpcm)

Dũng Nguyễn · Answer

Ta có:$\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2$

$=a^2+2ab+b^2+2ac+2bc+c^2$

$=a^2+b^2+c^2+2ab+2bc+2ca$Câu trả lời: Ta có:$\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2$

$=a^2+2ab+b^2+2ac+2bc+c^2$

$=a^2+b^2+c^2+2ab+2bc+2ca$

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