149/150

bạn giải rõ giúp mình được ko

Đặt A = 1x2+2x3+3x4+...+25x26+26x27

3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4  -2 x 3 x 4 + ..... + 26 x 27 x 28

3A = 26 x 27 x 28

A= \(\text{ }\frac{\text{26 x 27 x 28}}{3}=6552\)

Cậu học cấp 2 rồi còn giải bài cấp 1 làm chi.

giải chi tiết giúp mình

\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)

\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)

\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)

\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)

\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)

3C=1x2x3+2x3x3+3x4x3+...+Xx(X+1)=

=1x2x3+2x3x(4-1)+3x4x(5-2)+...+Xx(X+1)[(X+2)-(X-1)]=

=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-(X-1)xXx(X+1)+Xx(X+1)x(X+2)=

=Xx(X+1)(X+2)

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

A = 1 - \(\frac{1}{4}\)

A = \(\frac{3}{4}\)

A = 2 + 3/2 + 4/3

A = 2/1 + 3/2 + 4/3 

A = 12 / 6 + 9 / 6 + 8 / 6

A = 12/6 + ( 9/6 + 8/6 )

A = 12/6 + 17/6

A = 29/6

A = 1/2 + 1/6 + 1/12 

A = 6/12 + 2/12 + 1/12 

A =  9/12

A = 3/4

~ Hok ~

333300 nha bạn.Đúng thì tích cho mik với

1x2+2x3+...+19x20

3S= 1x2x3+2x3x3+3x4x3+...+19x20x3

3S=1x2x3+2x3x(4-1)+...+19x20x(21-3

3S=1x2x3+2x3x4-1x2x3+2x4x5-2x3x4+4x5x6+...19x20x21-18x19x20

S=19x20x21:3

S=7980

nhầm KQ=2660 nha

\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

1/1x2+1/2x3+1/3x4+...+1/999x1000

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000

=1-1/1000

=1000/1000-1/1000

=999/1000

999 / 1000

Gọi phép tính là: \(A\)

Ta có:

\(A=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{9\times10}\\ A\div2=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\\ A\div2=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ A\div2=\dfrac{9}{10}\\ A=\dfrac{18}{10}=\dfrac{9}{5}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=2\cdot\dfrac{9}{10}=\dfrac{9}{5}\)

Đặt phép tính  là A

\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\)

\(2A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(2A=1-\dfrac{1}{10}\)

\(2A=\dfrac{9}{10}\)

\(A=\dfrac{9}{5}\)