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5B = 5 +52 +53+....+52009
5B- B = 4B= (52010-1)
=> B= \(\frac{5^{2010}-1}{4}\)
**** CHO MÌNH NHA
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\(S=1+5+5^2+5^4+...+5^{200}\)
\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S-S=5^{202}-1\)
\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)
a) S = 1 + 52 + 54 + ... + 5200
=> 52S = 52.(1 + 52 + 54 + ... + 5200)
=> 25S = 52 + 54 + 56 + ... + 5202
=> 25S - S = (52 + 54 + 56 + ... + 5202) - (1 + 52 + 54 + ... + 5200)
=> 24S = 5202 - 1
=> S = \(\frac{5^{202}-1}{24}\)
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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
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a) \(A=4+4^2+4^3+...+4^{200}\)
\(4A=4^2+4^3+...+4^{201}\)
\(4A-A=3A=4^{201}-4\)
\(A=\frac{4^{201}-4}{3}\)
b) \(B=1+5+5^2+...+5^{2017}\)
\(5B=5+5^2+5^3+...+5^{2018}\)
\(5B-B=4B=5^{2018}-1\)
\(B=\frac{5^{2018}-1}{4}\)
c) \(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{500}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{499}}\)
\(3C-C=2C=1-\frac{1}{3^{500}}=\frac{3^{500}-1}{3^{500}}\)
\(C=\frac{\left(\frac{3^{500}-1}{3^{500}}\right)}{2}\)
T_i_c_k cho mình nha,có j ko hiểu cứ hỏi mình nhé ^^
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a: =>x+1/2=5
=>x=9/2
b: =>(x-1)^2=900
=>x-1=30 hoặc x-1=-30
=>x=-29 hoặc x=31
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Bài 2:
1: \(5^n+5^{n+2}=650\)
\(\Leftrightarrow5^n\cdot26=650\)
\(\Leftrightarrow5^n=25\)
hay x=2
2: \(32^{-n}\cdot16^n=1024\)
\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)
hay n=-10
13: \(9\cdot27^n=3^5\)
\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)
=>3n=3
hay n=1
5B = 52 + 53 + 54 + ... + 561
5B - B = (52 + 53 + 54 + ... + 561) - (5 + 52 + 53 + ... + 560)
4B = 561 - 5
\(B=\frac{5^{61}-5}{4}\)
Ta có: B = 5 + 52 + 53 + ... + 560
=> 5B = 52 + 53 + ... + 561
=> 5B - B = 561 - 5
=> 4B = 561 - 5
=> B = \(\frac{5^{61}-5}{4}\)