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2+2+2+2+2+2+2+2+2+2+2+..........+2 (235 số 2)=2.235=470
4+4+4+4+4+4+4+4+4+4+44+4=4.11+44=44.2=88
6+66+666+6666+66666=6+6.11+6.111+6.1111+6.11111=6.(1+11+111+1111+11111)=6.12345=74070
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
B=4+42+43+44+45+...+42017+42018
4B=42+43+44+45+46+...+42018+42019
\(\Rightarrow\)4B - B=42+43+44+45+46+...+42018+42019 - (4+42+43+44+45+..+42017+42018)
B=42+43+44+45+46+...+42018+42019 - 4-42-43-44-45-...-42017-42018
\(\Rightarrow\) B=42019-4
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\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{21.273.1333.4161.10101}{91.651.2451.6643.14763}\)
\(=\frac{3.7.13.21.31.43.73.57.91.111}{7.13.21.31.43.57.73.91.111.133}=\frac{3}{133}\)
Tuy nhiên cách làm trên phải có máy tính mới làm đc:
Có thể sử dụng công thức:
\(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Sau đó phân h:
\(2^4+2^2+1=\left(2^2+2+1\right)\left(2^2-2+1\right)=7.3\)
\(4^4+4^2+1=\left(4^2+4+1\right)\left(4^2-4+1\right)=21.13\)
....Tiếp tực làm thì sẽ ra đc kết quả:
\(=\frac{3.7.13.21.31.43.73.57.91.111}{7.13.21.31.43.57.73.91.111.133}=\frac{3}{133}\)
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`#3107.101107`
\(B=4+4^2+4^3+...+4^{89}+4^{90}\)
\(=\left(4+4^2+4^3\right)+...+\left(4^{88}+4^{89}+4^{90}\right)\)
\(=4\left(1+4+4^2\right)+...+4^{88}\left(1+4+4^2\right)\)
\(=\left(1+4+4^2\right)\left(4+...+4^{88}\right)\)
\(=21\left(4+4^{88}\right)\)
Vì \(21\left(4+4^{88}\right)\) `\vdots 21`
`\Rightarrow B \vdots 21`
Vậy, `B \vdots 21.`
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn \(D\)
Vì :
\(\dfrac{2}{4}=\dfrac{1}{2}\)
Còn đáp án \(A,B,C\) không đúng vì :
\(\dfrac{1}{4}=\dfrac{1}{4}\)
\(\dfrac{-1}{4}=\dfrac{-1}{4}\)
\(\dfrac{-2}{4}=\dfrac{-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2^{39}+2^{35}}{2^{33}\cdot4}\)
\(=\dfrac{2^{35}\cdot\left(2^4+1\right)}{2^{33}\cdot2^2}\)
\(=\dfrac{2^{35}\cdot\left(16+1\right)}{2^{35}}\)
\(=17\)
\(---\)
\(\dfrac{4^{17}+4^3}{4^{16}+4^2}\)
\(=\dfrac{4^3\cdot\left(4^{14}+1\right)}{4^2\cdot\left(4^{14}+1\right)}\)
\(=4\)
#\(Toru\)