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=> ĐK: \(x\ne\left\{0;-1;-2;...;-99;-100\right\}\)
Đây là dạng dãy số đặc biệt, bạn có thể giải như sau:
Ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{100}{x^2+100x}=\frac{100}{101}\)
\(\Leftrightarrow x^2+100x=101\)
\(\Leftrightarrow x^2+100x-101=0\)
\(\Leftrightarrow x^2+101x-x-101=0\)
\(\Leftrightarrow x\left(x+101\right)-\left(x+101\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+101\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+101=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(n\right)\\x=-101\left(n\right)\end{cases}}\)
Vậy: S={1;-101)
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\(\frac{\left(x+1\right)-x}{x\left(x+1\right)}+\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+...+\frac{\left(x+100\right)-\left(x+99\right)}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
Tự giải nha
![](https://rs.olm.vn/images/avt/0.png?1311)
- Với \(x=\left\{100;101\right\}\) là 2 nghiệm của pt
- Với \(x< 100\Rightarrow\left\{{}\begin{matrix}\left|x-100\right|>0\\\left|x-101\right|=\left|101-x\right|>1\end{matrix}\right.\)
\(\Rightarrow\left|x-100\right|^{100}+\left|x-101\right|^{101}>1\) ptvn
- Với \(x>101\Rightarrow\left\{{}\begin{matrix}\left|x-101\right|>0\\\left|x-100\right|>1\end{matrix}\right.\)
\(\Rightarrow\left|x-100\right|^{100}+\left|x-101\right|^{101}>1\) ptvn
- Với \(100< x< 101\Rightarrow\left\{{}\begin{matrix}0< x-100< 1\\0< 101-x< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-100\right|^{100}< x-100\\\left|x-101\right|^{101}=\left|101-x\right|^{101}< 101-x\end{matrix}\right.\)
\(\Rightarrow\left|x-100\right|^{100}+\left|x-101\right|^{101}< x-100+101-x=1\) ptvn
Vậy pt có đúng 2 nghiệm \(x=\left\{100;101\right\}\)
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\(\dfrac{x+101}{2001}+\dfrac{x+99}{2003}=\dfrac{x+100}{2002}+\dfrac{x+98}{2004}\)
\(\Leftrightarrow\left(\dfrac{x+101}{2001}+1\right)+\left(\dfrac{x+99}{2003}+1\right)=\left(\dfrac{x+100}{2002}+1\right)+\left(\dfrac{x+98}{2004}+1\right)\)
\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}=\dfrac{x+2102}{2002}+\dfrac{x+2102}{2004}\)
\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}-\dfrac{x+2102}{2002}-\dfrac{x+2102}{2004}=0\)
\(\Leftrightarrow\left(x+2102\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\right)=0\)
Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\ne0\)
\(\Rightarrow x+2102=0\)
\(\Rightarrow x=-2102\)
\(\Rightarrow S=\left\{-2102\right\}\)
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a) Ta có : $1.3+2.4+3.5+...+99.101+100.102$
$=(2-1)(2+1)+(3-1)(3+1)+(4-1)(4+1)+...+(100-1)(100+1)+(101-1)(101+1)$
$=2^2-1+3^2-1+4^2-1+...+100^2-1+101^2-1$
$=(2^2+3^2+4^2+...+100^2+101^2)-100$
b) $1.100+2.99+3.98+...+99.2+100.1$
$=1.100+2.(100-1)+3.(100-2)+...+99.(100-98)+100.(100-99)$
$=100(1+2+3+...+99+100)-(1.2+2.3+...+99.100)$
$=100.\dfrac{101.100}{2}-\dfrac{99.100.101}{3}=171700$