Ta có:

\(D=1.2+2.3+3.4+4.5+...+99.100\)

\(\Leftrightarrow3D=1.2.\left(3-0\right)+2.3+\left(4-1\right)+3.4+\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)

\(\Leftrightarrow3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)

\(\Leftrightarrow3D=99.100.101\Leftrightarrow D=\frac{99.100.101}{3}=333300\)

\(B=1.3+2.4+3.5+4.6+...+99.101\)

\(\Leftrightarrow B=\left(1.3+3.5+...+99.101\right)+\left(2.4+4.6+...+98.100\right)\)

\(\Leftrightarrow6B=\left(1.3.\left(5-\left(-1\right)\right)+3.5.\left(7-1\right)+...+99.101.\left(103-97\right)\right)+\left(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\right)\)

\(\Leftrightarrow B=\frac{99.101.103+3}{6}+\frac{98.100.102}{6}=338250\)

Vì các bước gần tương tự như bài a) nên mình bỏ bước.

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)

1) 5(x - 2) + 27 = 4x - 8

=> 5x - 10 + 27 = 4x - 8

=> 5x + 17 = 4x - 8

=> 5x - 4x = -8 - 17

=> x = -25

2) 3(x + 1) + 2(x  + 2) = -13

=> 3x + 3 + 2x + 4 = -13

=> 5x + 7 = -13

=> 5x = -13 - 7

=> 5x = -20

=> x = -20 : 5

=> x = -4

3) (2x + 4)(5 - x) = 0

=> \(\orbr{\begin{cases}2x+4=0\\5-x=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=-4\\x=5\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=5\end{cases}}\)

Vậy ...

4) x² - 3x = 0

=> x(x - 3) = 0

=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy ...

5) x² - 3x + 2 = 0

=> x(x - 3) = -2

=> x(x - 3)  = 1 . (1 - 3)

=> x = 1

\(5\frac{9}{10}:\frac{3}{2}-\left[2\frac{1}{3}\times4\frac{1}{2}-2\times2\frac{1}{3}\right]:\frac{7}{4}\)

\(=\frac{59}{10}:\frac{3}{2}-\left[\frac{7}{3}\times\frac{9}{2}-2\times\frac{7}{3}\right]:\frac{7}{4}\)

\(=\frac{59}{10}\times\frac{2}{3}-\left[\left(\frac{9}{2}-2\right)\times\frac{7}{3}\right]:\frac{7}{4}\)

\(=\frac{59}{10}\times\frac{2}{3}-\left[\frac{5}{2}\times\frac{7}{3}\right]:\frac{7}{4}\)

\(=\frac{59}{10}\times\frac{2}{3}-\frac{35}{6}:\frac{7}{4}\)

\(=\frac{59}{15}-\frac{35}{6}\times\frac{4}{7}\)

\(=\frac{59}{15}-\frac{221}{42}\)

\(=-\frac{93}{70}\)

Học tốt !

- Vc Lớp 5 

32,2857 và 12,8 và -498814315,3 và 0,1 và 11 

tổng = - 498814259,1

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{21}.\left(1+2+3+...+21\right)\)

\(A=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{21}.\frac{\left(1+21\right).21}{2}\)

\(A=1+\frac{3}{2}+\frac{4}{2}+...+\frac{22}{2}\)

\(A=\left(1+11\right).21:2=231\)

sorry, đáp án của mình là 126 nha bạn

đáp số = -3/4 đấy bạn

S=2/2+3/2+4/2+...+102/2
S=(2+102)*51/2