Ta có: C=32+(32)2+(32)3+....+(32)2013⇔32C=32.[32+(32)2+(32)3+....+(32)2013]⇔32C=(32)2+(32)3+(32)4+........

Ta có:C=32+(32)2+(32)3+....+(32)2013&#x2...

trinh anh tan

5 tháng 9 2020 - olm

Ta có:

$C = 32 + (32) 2 + (32) 3 + . . . . + (32) 2013 ⇔ 32 C = 32 . [32 + (32) 2 + (32) 3 + . . . . + (32) 2013] ⇔ 32 C = (32) 2 + (32) 3 + (32) 4 + . . . . . . + (32) 2014 ⇔ 32 C − C = [(32) 2 + (32) 3 + (32) 4 + . . . . . . + (32) 2014] − [32 + (32) 2 + (32) 3 + . . . . + (32) 2013] ⇔ 12 C = (32) 2014 − (32) ⇔ C = (32) 2014 .2 − 3 A = 12 + C = (32) 2014 .2 − 52 B − A = ((32) 2013 .2) − ((32) 2014 .2 − 52) = 2. (32) 2013 . (1 − 32) + 52 = − (32) 2013 + 52 " role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:17.55px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; text-align:left; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml"> \begin{matrix} C = \frac{3}{2} + {(\frac{3}{2})}^{2} + {(\frac{3}{2})}^{3} + . . . . + {(\frac{3}{2})}^{2013} \Leftrightarrow \frac{3}{2} C = \frac{3}{2} . [\frac{3}{2} + {(\frac{3}{2})}^{2} + {(\frac{3}{2})}^{3} + . . . . + {(\frac{3}{2})}^{2013}] \Leftrightarrow \frac{3}{2} C = {(\frac{3}{2})}^{2} + {(\frac{3}{2})}^{3} + {(\frac{3}{2})}^{4} + . . . . . . + {(\frac{3}{2})}^{2014} \Leftrightarrow \frac{3}{2} C - C = [{(\frac{3}{2})}^{2} + {(\frac{3}{2})}^{3} + {(\frac{3}{2})}^{4} + . . . . . . + {(\frac{3}{2})}^{2014}] - [\frac{3}{2} + {(\frac{3}{2})}^{2} + {(\frac{3}{2})}^{3} + . . . . + {(\frac{3}{2})}^{2013}] \Leftrightarrow \frac{1}{2} C = {(\frac{3}{2})}^{2014} - (\frac{3}{2}) \Leftrightarrow C = {(\frac{3}{2})}^{2014} .2 - 3 A = \frac{1}{2} + C = {(\frac{3}{2})}^{2014} .2 - \frac{5}{2} B - A = ({(\frac{3}{2})}^{2013} .2) - ({(\frac{3}{2})}^{2014} .2 - \frac{5}{2}) = 2. {(\frac{3}{2})}^{2013} . (1 - \frac{3}{2}) + \frac{5}{2} = - {(\frac{3}{2})}^{2013} + \frac{5}{2} \end{matrix}$

#Hỏi cộng đồng OLM #Toán lớp 6

tìm x ∈ Z biết : 16( $x - 1$ $x - 1$ = (x − $y - 1$ $y - 1$

tìm x ; y biết : 3.x $x$ $x$ .y + 5. $x$ $x$ $x$ $y$ $y$ + 5. $y$
$y$ = 16

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