Tìm x biết\(\left|x-1\right|+\left|x-2\right|+.....+\left|x-100\right|=101\)
tìm x biết: \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)
Vì \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\forall x\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\forall x\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
Từ điều kiện trên ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(100x+\frac{1+2+...+100}{101}=101x\)
\(101x-100x=\frac{5050}{101}\)
\(x=50\)
Vậy x = 50
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)
\(KĐ:101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)
\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)
\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)
\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)
\(x=\frac{101.100:2}{101}\)
\(x=50\)
5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
tìm x biết
a)\(x+2x+3x+4x+...+2015x=2016\times2017\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1008}-1}{4}\)
c)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x\)
d)tìm x nguyên biết \(\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|=2500\)
e) tìm x nguyên biết \(2004=\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+99x\right|+\left|x+1000\right|\)
tìm x biết: \(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+....+\left|x+\dfrac{100}{101}\right|=101x\)
\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=101x\)
Ta có : \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|\ge0\\\left|x+\dfrac{1}{102}\right|\ge0\\....\\\left|x+\dfrac{100}{101}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{101}\right|=x+\dfrac{1}{101}\\\left|x+\dfrac{2}{101}\right|=x+\dfrac{2}{101}\\....\\\left|x+\dfrac{100}{101}\right|=x+\dfrac{100}{101}\end{matrix}\right.\)
\(\Rightarrow x+\dfrac{1}{101}+x+\dfrac{2}{101}+x+\dfrac{3}{101}+...+x+\dfrac{100}{101}=101x\)
\(\Rightarrow100x+\dfrac{1+2+3+...+100}{101}=101x\)
\(\Rightarrow100x+\dfrac{5050}{101}=101x\)
\(\Rightarrow100x+50=101x\)
\(\Rightarrow101x-100x=50\)
\(\Rightarrow x=50\)
Vậy \(x=50\)
Câu 1: Tìm x biết:
a)\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
b)\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
c)\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
d)\(\left|x+\frac{1}{1.5}\right|+\left|x+\frac{1}{5.9}\right|+\left|x+\frac{1}{9.13}\right|+...+\left|x+\frac{1}{397.401}\right|=101x\)
Nhận xét :
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
b)
Tương tự câu a) , phương trình tương đương với :
\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)
\(\Rightarrow x=\frac{97}{195}\)
c)
Tương tự câu a) , phương trình tương đương với :
\(99x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)
\(\Rightarrow x=\frac{99}{100}\)
Tìm x, biết
\(\frac{1}{x\left(x+1\right)}\)+ \(\frac{1}{\left(x+1\right)\left(x+2\right)}\)+ ... + \(\frac{1}{\left(x+99\right)\left(x+100\right)}\)= \(\frac{100}{101}\)
=> 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)
Tìm x
a) \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+.....+\left(x-100\right)=101\)
b) \(x+2x+3x+....+100x=1000\)
c) \(x+1+2\left(x+2\right)+3\left(x+3\right)+...+100\left(x+100\right)=101^2\)
d) \(\frac{1+x}{1}+\frac{1+x}{2}+\frac{1+x}{3}+...+\frac{1+x}{30}=0\)
e) \(\left(1+\frac{x}{1}\right)\left(2-\frac{x}{2}\right)\left(3-\frac{x}{3}\right)=0\)
BẠN NÀO BIẾT PHẦN NÀO THÌ GIÚP MIK NHÉ!
Thank!!
a) (x-1)+(x-2)+(x-3)+...+(-100)=101
(x+x+x+...+x)-(1+2+3+...+100)=101
=> 100x-5050=101
100x=101+5050
100x=5151
x=5151:100
x=5151/100
Tìm \(x\in Q\) sao cho:
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{1}{102}\right|+....+\left|x+\frac{100}{101}\right|>0\)
\(\Rightarrow101x>0\)
\(\Rightarrow x>0\)
\(\Rightarrow\left(x+\frac{1}{101}\right)+.....+\left(x+\frac{100}{101}\right)=101x\)
\(\Rightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)
\(\Rightarrow x=\frac{\left(100+1\right)100:2}{101}\)
\(\Rightarrow x=\frac{50.101}{101}\)
\(\Rightarrow x=50\)
Vậy x = 50
Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
=> \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x 100 phân số
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Tìm x, biết:
\(\frac{1}{x\left(x+1\right)}\)+ \(\frac{1}{\left(x+1\right)\left(x+2\right)}\)+ ... + \(\frac{1}{\left(x+99\right)\left(x+100\right)}\)= \(\frac{100}{101}\)
\(\frac{\left(x+1\right)-x}{x\left(x+1\right)}+\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+...+\frac{\left(x+100\right)-\left(x+99\right)}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
Tự giải nha
\(\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{100}{x\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow x\left(x+100\right)=101\)
\(\orbr{\begin{cases}x=1\\x=-101\end{cases}}\)
\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+\left|x+\dfrac{3}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=101x\)
\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)
VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)
Ta có: \(\left|x+\frac{1}{101}\right|\ge0\) \(\forall x\)
\(\left|x+\frac{2}{101}\right|\ge0\) \(\forall x\)
\(\left|x+\frac{3}{101}\right|\ge0\) \(\forall x\)
\(............\)
\(\left|x+\frac{100}{101}\right|\ge0\) ∀\(x\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\) \(\forall x\)
\(\Leftrightarrow101x\ge0\)
\(\Leftrightarrow x\ge0\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 hạng tử x 100 số hạng
\(\Leftrightarrow100x+\left(\frac{\left(100+1\right)\cdot100:2}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{101\cdot50}{101}=101x\)
\(\Leftrightarrow50=101x-100x\)
\(\Rightarrow x=50\)