A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)

2\(\times\)A =  1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)

2\(\times\)A - A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\))

A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\) - \(\dfrac{1}{64}\)

A = (1 - \(\dfrac{1}{64}\))+(\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\))+(\(\dfrac{1}{4}\) - \(\dfrac{1}{4}\))+(\(\dfrac{1}{8}\)-\(\dfrac{1}{8}\))+(\(\dfrac{1}{16}\) - \(\dfrac{1}{16}\))+(\(\dfrac{1}{32}\)-\(\dfrac{1}{32}\))

A = 1 - \(\dfrac{1}{64}\) 

A = \(\dfrac{63}{64}\)

12+14+18+116+132+164

=1−12+12−14+14−18+18−116+116−132+132−164=1−21​+21​−41​+41​−81​+81​−161​+161​−321​+321​−641​

=1−164=1−641​

=6364=6463​
 

giảm chiều dài đi 6 m hay bao nhiêu vậy em?

tăng chiều rộng 6m và giảm chiều dài đi 6m cô ạ 

2²⁰²⁰ hay 2²⁰²² em ơi?

Hạng tử cuối ý em.

khó vậy

số học sinh thích môn toán là:

40:100x30=12(hs)

số học sinh thích môn toán là:

40:100x30=12   hs

a: y=ax+b(a>0)

b: y=ax+b(a<0)

\(\left\{{}\begin{matrix}45\left(t+\dfrac{1}{2}\right)=S\\60\left(t-\dfrac{3}{4}\right)=S\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}45\left(t+\dfrac{1}{2}\right)=60\left(t-\dfrac{3}{4}\right)\\S=45\left(t+\dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left(t+\dfrac{1}{2}\right)=4\left(t-\dfrac{3}{4}\right)\\S=45\left(t+\dfrac{1}{2}\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3t+1,5=4t-3\\S=45\left(t+0,5\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-t=-4,5\\S=45\left(t+0,5\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}t=4,5\\S=45\left(4,5+0,5\right)=45\cdot5=225\end{matrix}\right.\)