Câu hỏi của Hòa Đình

Question

rút gọn biểu thức

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

b , $\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)$

Trần Việt Linh · Accepted Answer

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

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

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

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

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

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

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

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

$=a^2$