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\(C=1.2.3+2.3.4+...+48.49.50\)
\(\Rightarrow4C=1.2.3.4+2.3.4.4+...+48.49.50.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+...+48.49.50.51-47.48.49.50\)
\(=48.49.50.51\)
\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1.2.3+2.3.4+...+98.99.100.\)
\(\Rightarrow4S=4\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(=1.2.3.4+2.3.4.4+...+98.99.100.4\)
\(=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+98.99.100.\left(101-97\right)\)
\(=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+98.99.100.101-97.98.99.100\)
\(=98.99.100.101\)
\(\Rightarrow S=\frac{98.99.100.101}{4}=24497550\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
Nếu đúng thì k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 1+1+1+....+1 ( có 50 số 1 )
= 1 x 50 = 50
Vậy B = 50
k mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
ta co:2x^2-2xy=5x-y-19 | |||||||||||||||||||||||||||||||||||
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Giải
Ta có: \(\left(2x+1\right)\left(y^2-5\right)=12\)
\(\Leftrightarrow\hept{\begin{cases}2x+1\\y^2-5\end{cases}}\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm4;\pm6;\pm3;\pm12\right\}\)
Lập bảng:
\(2x+1\) | \(-1\) | \(-2\) | \(-3\) | \(-4\) | \(-6\) | \(-12\) | \(1\) | \(2\) | \(3\) | \(4\) | \(6\) | \(12\) |
\(y^2-5\) | \(-12\) | \(-6\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(12\) | \(6\) | \(4\) | \(3\) | \(2\) | \(1\) |
\(x\) | \(-1\) | Loại | \(-2\) | Loại | \(1\) | |||||||
\(y\) | Loại | Loại | Loại | Loại | Loại | Loại | Loại | Loại | \(3\) | Loại | Loại | Loại |
Vậy x =1 và y = 3
Ta có :
\(\text{A = 1.2.3 + 3.4.5+...99.100.101}\)
\(\text{A=1.3(5-3)+3.5(7-3)+}...+99.101\left(103-3\right)\)
\(=\left(1.3.5+3.5.7+5.7.9+...99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)
\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+...+99.101\right)\)
\(=13517400-3.171650\)
\(=13002450\)
D=1.2.3+3.4.5+...+99.100.101
D=1.2.3.4+5.6.7.4+........+99.100.101.4
D=1.2.3.4+5.6.7.(8-4)+........+99.100.101.(102-98)
D=(1.2.3.4+5.6.7.8+.........+99.100.101.102)-(1.2.3.4+5.6.7.8+....+98.99.100.101)
D=98.99.100.101