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( \(1+\frac{1}{x}\))(\(1+\frac{1}{x+1}\))(\(1+\frac{1}{x+2}\))....(\(1+\frac{1}{x+99}\))
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PV
16 tháng 3 2017
1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )
= 29 x 6 - 19 x 16
= 174 - 304
= - 130
2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= 1 - \(\frac{1}{100}\)
= \(\frac{99}{100}\)
LJ
9 tháng 11 2017
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+.....+\frac{1}{x+99}-\frac{1}{x+100}\)
\(=\frac{1}{x}-\frac{1}{x+100}\)
BT
9 tháng 11 2017
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{x\left(x+100\right)}\)
PA
1 tháng 12 2016
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right)\times\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left[\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right]\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x^2-x+1\right)-3+3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3\left(x+1\right)^2}{\left(x+1\right)\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3x}{x\left(x+2\right)}-\frac{2x-2}{x\left(x+2\right)}\)
\(=\frac{3x-2x+2}{x\left(x+2\right)}\)
\(=\frac{x+2}{x\left(x+2\right)}\)
\(=\frac{1}{x}\)
13 tháng 10 2019
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)
NC
14 tháng 8 2020
Xin phép sửa đề:
Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)
\(\Leftrightarrow\frac{\left(3x+1\right)\left(x+1\right)-\left(1-x\right)^2}{\left(1-x\right)^2\left(x+1\right)}=\frac{\left(x+3\right)\left(1-x\right)}{\left(1-x\right)^2\left(x+1\right)}\)
\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)
\(\Leftrightarrow3x^2+8x-3=0\)
\(\Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\)
\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)
\(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right)...\left(1+\frac{1}{x+99}\right)\)
\(=\frac{x+1}{x}.\frac{x+2}{x+1}.\frac{x+3}{x+2}...\frac{x+100}{x+99}=\frac{\left(x+1\right)\left(x+2\right)\left(x+3\right)...\left(x+100\right)}{x\left(x+1\right)\left(x+2\right)...\left(x+99\right)}=\frac{x+100}{x}\)