1*2^2*2^4*.......*2^98 giúp mình đi mình đg cần gấp
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\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)
\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)
\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).
\(A=2^2+4^4+6^2+...98^2\)
\(\frac{1}{4}A=1^2+2^2+3^2+...+49^2\)
\(\frac{1}{4}A=\frac{49.50.\left(2.49+1\right)}{6}\)
\(\frac{1}{4}A=40425\)
\(A=40425.4=161700\)
=>\(2^2+4^2+6^2+...+98^2=161700\)
(2xy-4/9)3/2=6/11
<=>3xy-12/18=6/11
<=>99xy/33-22/33=18/33
<=>99xy-22=18
<=>99xy=18+22
<=>99xy=40
<=>xy=40/99
(2xy-4/9)3/2=6/11
<=>3xy-12/18=6/11
<=>99xy/33-22/33=18/33
<=>99xy-22=18
<=>99xy=18+22
<=>99xy=40
<=>xy=40/99
=1/4 + 11/10 : (2/5 - 3/2) - 1/9
=1/4 + 11/10 : (4/10 - 15/10) - 1/9
=1/4 + 11/10 : -11/10 - 1/9
=1/4 + (-1) - 1/9
=1/4 + (-4/4) - 1/9
=-3/4 + 1/9
=-27/36 + 4/36
=-23/36
1*2^2*2^4*.......*2^98 =2^0.2^2.2^4.....2^98=2^0+2+4+....+98(vì 1=2^0)
mà 0+2+....+98=[(98-2)]:2+1.(98+2):2=2500
=> 1*2^2*2^4*.......*2^98 =2^2500