4:3=?
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+3^4+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+3^5+...+3^{51}\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+4^4+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+4^5+...+4^{2019}\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)
\(B=3+3^2+3^3+...+3^{50}\)
\(\Rightarrow3B=3^2+3^3+3^4+....+3^{51}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{51}\right)-\left(3+3^2+...+3^{50}\right)\)
\(\Rightarrow2B=3^{51}-3\)
\(\Rightarrow B=\frac{3^{51}-3}{2}\)
\(C=4+4^2+4^3+...+4^{2018}\)
\(\Rightarrow4C=4^2+4^3+4^4+....+4^{2019}\)
\(\Rightarrow4C-C=\left(4^2+4^3+4^4+...+4^{2019}\right)-\left(4+4^2+4^3+...+4^{2018}\right)\)
\(\Rightarrow3C=4^{2019}-4\)
\(\Rightarrow C=\frac{4^{2019}-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
xét B=-3/4+(3/4)^2-.......-(3/4)^n với n lẻ,n>=1
=>-3/4.B=(3/4)^2-(3/4)^3+.........+(3/4)...
trừ theo vế suy ra 7/4.B=-3/4-(3/4)^(n+1)
=>7B=-3-(3/4)^n
=>A=1+B=1-(3+(3/4)^n)/7
do <0(3/4)^n <1
suy ra 0< 3+(3/4)^n <7
suy ra (3+(3/4)^n)/7 ko là số nguyên
suy ra A ko nguyên
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=-3\left(\dfrac{1}{4}-\dfrac{1}{4^2}+\dfrac{1}{4^3}-\dfrac{1}{4^4}+...-\dfrac{1}{4^{100}}\right)\)
Đặt \(C=\dfrac{1}{4}-\dfrac{1}{4^2}+...-\dfrac{1}{4^{100}}\)
\(\Leftrightarrow C\cdot\dfrac{1}{4}=\dfrac{1}{4^2}-\dfrac{1}{4^3}+...-\dfrac{1}{4^{101}}\)
\(\Leftrightarrow C\cdot\dfrac{-3}{4}=\dfrac{-1}{4^{101}}-\dfrac{1}{4}=\dfrac{-1-4^{100}}{4^{101}}\)
\(\Leftrightarrow C=\dfrac{-4^{100}-1}{4^{101}}\cdot\dfrac{-4}{3}=\dfrac{4^{100}+1}{3\cdot4^{100}}\)
\(\Leftrightarrow B=\dfrac{-4^{100}-1}{4^{100}}\)
4:tứ
3:tam
tứ chia tam bằng tám chia tư bằng 2
C1) 4:3=4/3
C2) 4:3=tứ :tam=8:4=2 k cho mình nha