Hằng đẳng thức đáng nhớ(Rút gọn)

E

eunhyuk_0330

Câu b:
$B=(2+1)(2^2+1)(2^4+1)(2^8+1)$
\Leftrightarrow $B(2-1)=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)$
\Leftrightarrow$B=(2^2-1)(2^2+1)(2^4+1)(2^8+1)$
$=(2^4-1)(2^4+1)(2^8+1)$
$=(2^8-1)(2^8+1)$
$=2^{16}-1$
 
D

depvazoi

$A=100^2-99^2+98^2-97^2+...+2^2-1^2$
$=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)$
$=100+99+98+97+...+2+1$
$=\dfrac{(100+1).100}{2}$
$=5050$
 
M

manhngughe

A=1002−992+982−972+...+22−12
=(100−99)(100+99)+(98−97)(98+97)+...+(2−1)(2+1)
=100+99+98+97+...+2+1
=(100+1).1002
=5050

Câu b:
B=(2+1)(22+1)(24+1)(28+1)
B(2−1)=(2−1)(2+1)(22+1)(24+1)(28+1)
B=(22−1)(22+1)(24+1)(28+1)
=(24−1)(24+1)(28+1)
=(28−1)(28+1)
=216−1
Trả Lời Với Trích Dẫn
 
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