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=> 52S=52+54+56+...+5202
=>52S-S= (52+54+56+...+5202)-(1+52+54+...+5200)
=> 24.S = 5202-1
=> S = \(\frac{5^{202}-1}{24}\)
\(3^{-200}=\left(3^{-2}\right)^{100}=\left(\frac{1}{9}\right)^{100}\)
\(2^{-300}=\left(2^{-3}\right)^{100}=\left(\frac{1}{8}\right)^{100}\)
\(\frac{1}{9}< \frac{1}{8}\Rightarrow\left(\frac{1}{9}\right)^{100}< \left(\frac{1}{8}\right)^{100}\Rightarrow3^{-200}< 2^{-300}\)
\(33^{52}=\left(33^4\right)^{13}\)
\(44^{39}=\left(44^3\right)^{13}\)
\(33^4=\left(33^{\frac{4}{3}}\right)^3\approx106^3\)
\(106^3>44^3\Rightarrow\left(33^4\right)^{13}> \left(44^3\right)^{13}\Rightarrow33^{52}>44^{39}\)
a) Ta có: 2300=(23)100=8100
3200=(32)100=9100
Vì 8<9 nên 8100<9100
Vậy 2300<3200
b) Ta có: 2333=(23)111=8111
3222=(32)111=9111
Vì 8<9 nên 8111<9111
Vậy 2333<3222
a) 2300 = 23 . 100 = ( 23 )100 = 8100
3200 = 32 . 100 = ( 32 )100 = 9100
Vì 8100 < 9100 nên 2300 < 3200
b) Tương tự
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
a/ P=1-1/2+1/3-1/4+....+1/199-1/200
= 1+1/2+1/3+1/4+1/5+...+1/200 - 2.(1/2+1/4+...+1/200)
= 1+1/2+1/3+1/4+1/5+...+1/200 - 1-1/2-1/3-...-1/100
=1/101+1/102+...+1/200
b/ k-k/2+ k/3- k/4+...+k/199-k/200
=k+k/2+k/2+...+k/199+k/200 -2(k/2+k/4+k/6+...+k/200)
=k+k/2+k/2+...+k/199+k/200-k-k/2-k/3-...-k/100
=k/101+k/102+...+k.200
a) 2100 = ( 22 )50 = 450
Ta có : 450 > 350
=> 2100 > 350 ( đpcm )
b) đương nhiên là 3200 lớn hơn rồi.
c) Ta có : 5222 = ( 52 )111 = 25111 ( 1)
2555 = ( 25 )111 = 32111 ( 2)
Từ (1) và (2) => 5222<2555