Câu hỏi của Lê Hải Yến

Question

Rút gọn:

a. (2a+b)^2-(2a+b)(2a-b)-2a(b-a).

b. (a+b-c)^2-(a+b)^2+2c(a+b).

Khôi Bùi · Accepted Answer

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

$=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2$

$=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2$

$=2a^2+2ab+2b^2$

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

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

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

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

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

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

$=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2$

$=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2$

$=2a^2+2ab+2b^2$

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

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

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

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

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

$=c^2$

Thư Vy · Answer

@Khôi Bùi

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

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

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

$=4ab+2b^2-2ab+2a^2=2\left(a^2+ab+b^2\right)$

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

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

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

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