Câu hỏi của Shu Sakamaki

Question

Tính: $\left(a+b+c\right)^2$$\left(a-b-c\right)^2$Mọi người giúp mk với ạ. Mk đang cần gấp.

Aikawa Maiya · Accepted Answer

$\left(a+b+c\right)^2$$\Rightarrow\left[\left(a+b\right)+c\right]^2$$\Rightarrow\left(a+b\right)^2+2c\left(a+b\right)+c^2$$\Rightarrow a^2+2ab+b^2+2ca+2bc+c^2$$\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca$$\left(a-b-c\right)^2$$\Rightarrow\left[\left(a-b\right)-c\right]^2$$\Rightarrow\left(a-b\right)^2-2c\left(a-b\right)+c^2$$\Rightarrow a^2-2ab+b^2-2ca+2bc+c^2$$\Rightarrow a^2+b^2+c^2-2ab+2bc-2ca$Câu trả lời: $\left(a+b+c\right)^2$$\Rightarrow\left[\left(a+b\right)+c\right]^2$$\Rightarrow\left(a+b\right)^2+2c\left(a+b\right)+c^2$$\Rightarrow a^2+2ab+b^2+2ca+2bc+c^2$$\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca$$\left(a-b-c\right)^2$$\Rightarrow\left[\left(a-b\right)-c\right]^2$$\Rightarrow\left(a-b\right)^2-2c\left(a-b\right)+c^2$$\Rightarrow a^2-2ab+b^2-2ca+2bc+c^2$$\Rightarrow a^2+b^2+c^2-2ab+2bc-2ca$

Thanh Trang · Answer

ta có (a+b+c)^2 = (a+b+c).(a+b+c) =a^2+ab+ac+ab+b^2+bc+ac+bc+c^2 = a^2+b^2+c^2+2ab+2ac+2bc   và   (a-b-c)^2 = (a-b-c)(a-b-c)     = a^2-ab-ac-(ab-b^2-bc)-(ac-cb-c^2)   =a^2-ab-ac-ab+b^2+bc-ac+cb+c^2=a^2 -2ab-2ac+bc+b^2+c^2Câu trả lời: ta có (a+b+c)^2 = (a+b+c).(a+b+c) =a^2+ab+ac+ab+b^2+bc+ac+bc+c^2 = a^2+b^2+c^2+2ab+2ac+2bc   và   (a-b-c)^2 = (a-b-c)(a-b-c)     = a^2-ab-ac-(ab-b^2-bc)-(ac-cb-c^2)   =a^2-ab-ac-ab+b^2+bc-ac+cb+c^2=a^2 -2ab-2ac+bc+b^2+c^2

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