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.

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

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

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

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

Câu 1 :

\(x:\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{101.103}\right)=1\)

\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\right]\) \(=1\)

\(=>x:\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{103}\right)\right]=1\)

\(=>\) \(x:\frac{51}{103}=1\)

\(=>x=1.\frac{51}{103}=\frac{51}{103}\)

Câu 2 :

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\right).x=2\)

\(=>\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right).x=2\)

\(=>\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)

\(=>\frac{11}{12}.x=2\)

\(=>x=2:\frac{11}{12}\)

\(=>x=\frac{24}{11}\)

cau b bang 24/11​

​

Cau a bang 51/103

​

1-1/x+1=2015/2016

=>1/x+1=1-2015/2016=1/2016

=>x+1=2016=>x=2015

mình không ghi lại đề nha:

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

<=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

<=>\(\frac{x}{x+1}=\frac{2015}{2016}\)

=>x=

Đến đó bạn tự giải tiếp ha

=>(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/x-1/x+1)=2015/2016
=>1/1-1/x+1=2015/2016
=>x/x+1=2015/2016
=>x=2015

a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8

=1.2.3.4...8(9-1-8)

=1.2.3.4...8.0

=0

b)(3.4.2¹⁶)²/11.12³.4¹¹-16⁹=(3.2².2¹⁶)²/11.2¹³.2²²-2³⁶=3².2⁴.2³²/11.2³⁵-2³⁶=3².2²⁶/2³⁵.(11-2)

=3².2³⁶/2³⁵.9=3².2³⁶/2³⁵.3²=2

c)70.(131313/565656+131313/727272+131313/909090

=70.(13/56+13/72+13/90)

=70.39/70=39

d)1/4.9+1/9.14+1/14.19+...+1/64.69

=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.

=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)

=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)

=1/4.(1/4-1/69)

=1/4.65/276=65/1104

~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~

nhìu thế này sao mà làm nổi

d)\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\right)\)

\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{69}\right)\)

\(\frac{1}{3}\times\frac{65}{276}\)

\(A=\frac{1^2}{1\times2}\times\frac{2^2}{2\times3}\times\frac{3^2}{3\times4}\times\frac{4^2}{4\times5}\)

      \(=\frac{1}{2}\times\frac{4}{6}\times\frac{9}{12}\times\frac{16}{20}\)

      \(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)

Gạch các số giống nhau của phép nhân đó là 2; 3; 4. Ta được kết quả bằng

      \(=\frac{1}{5}\)

Kết quả là   \(\frac{1}{5}\)

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)

\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)

\(\frac{2009}{2010}\cdot x=2009\)

\(x=2009:\frac{2009}{2010}\)

\(x=2010\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+....+\frac{1}{2009}-\frac{1}{2010}\right).x=2009\)

\(\left(\frac{1}{1}-\frac{1}{2010}\right).x=2009\)

\(\frac{2009}{2010}.x=2009\)

\(x=2009:\frac{2009}{2010}\)

\(x=2010\)

x = 2010

\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)

\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)

\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)

Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)

         = \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

        = 1 - \(\frac{1}{10}\)

        =  \(\frac{9}{10}\)

X = \(\frac{3}{2}\) là đúng đó bạn