1x2+2x3+3x4+...+x(x+1)
giúp mih với
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\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)
\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)
\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)
\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)
\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)
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)
`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`
`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`
`=>x/(x+1)=1-1/32`
`=>x/(x+1)=31/32`
`=>32x=31(x+1)`
`=>32x=31x+31`
`=>32x-31x=31`
`=>x=31`
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)
1/1.2 +1/2.3 +1/3.4 +....+1/99.100
=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100
=1-1/100
=99/100
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
Ta có : A = \(\frac{1}{1\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{3\text{x}4}+...+\frac{1}{X\text{x}\left(X+1\right)}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
A = \(\frac{1}{1}-\frac{1}{x+1}\)
A = \(\frac{x}{x+1}\)
Ủng hộ mik nhá !!!!
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-?\)
\(\Rightarrow x+1=?\Leftrightarrow x=?\)
Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này?
tih bieu thuc ma