Câu hỏi của Kirito1962005

Question

rút gọn $\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)$

Dũng Nguyễn · Accepted Answer

ta có: $\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)$

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

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

$=c\left(2a+2b-2a-2b+c\right)=c^2$Câu trả lời: ta có: $\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)$

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

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

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

Sáng · Answer

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

$=a^2+2ab+b^2-2bc+c^2-2ac-\left(a^2+2ab+b^2\right)+2ac+2bc$

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

$=c^2$Câu trả lời: $\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)$

$=a^2+2ab+b^2-2bc+c^2-2ac-\left(a^2+2ab+b^2\right)+2ac+2bc$

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

$=c^2$

Bài 3: Những hằng đẳng thức đáng nhớ