3+3+3*4=?
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Answer:
Chứng tỏ không phải số nguyên nhỉ?
\(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)
\(\Rightarrow A.\frac{3}{4}=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2010}+\left(\frac{3}{4}\right)^{2011}\)
\(\Rightarrow\frac{3}{4}A+A=\left(\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2010}+\left(\frac{3}{4}\right)^{2011}\right)+\left(1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\right)\)
\(\Rightarrow\frac{7}{4}A=\left(\frac{3}{4}\right)^{2011}+1\)
\(\Rightarrow A=\frac{4.\left(\frac{3}{4}\right)^{2011}+4}{7}\)
Vậy A không phải số nguyên
A = 1 - (3/4) + (3/4)² - (3/4)³ + ... - (3/4)^2009 + (3/4)^2010
A.(3/4) = (3/4) - (3/4)² + (3/4)³ - (3/4)^4 +... - (3/4)^2010 + (3/4)^2011
cộng 2 đẳng thức trên lại vế theo vế:
A + A.(3/4) = 1 + (3/4)^2011 => 7A/4 = 1 + (3/4)^2011
=> 7A = 4 + 4.(3/4)^2011 không là số nguyên => A không nguyên
vậy A ko phải là số nguyên
\(B=3+3^2+3^3+3^4+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+3^5+...+3^{51}\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+4^4+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+4^5+...+4^{2019}\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)
\(B=3+3^2+3^3+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+....+3^{51}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{51}\right)-\left(3+3^2+...+3^{50}\right)\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+....+4^{2019}\)
\(\Rightarrow4C-C=\left(4^2+4^3+4^4+...+4^{2019}\right)-\left(4+4^2+4^3+...+4^{2018}\right)\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)
1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
3+3+3*4=18
=3+3+12
=> 6+12=18
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