\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2005\right|=2006x\)
Xét: \(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
... \(\left|x+2005\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+2005\right|\)
= (x +1) +(x+2) +...+ (x+2005)
= x+1 +x+2+...+ x+2005
= (x+ x+ x+...+x)+ (1+2+...+2005)
= 2005x+ 2011015= 2006x
=> x= 2011015
Vậy x= 2011015
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x=\frac{1}{20}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
2. Tìm x, y, z biết\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
3.Tìm x\(a,2009-\left|x-2009\right|=x\)
\(b,\left|3x+2\right|=\left|5x-3\right|\)
Tim x biet :
\(a,\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(b,\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
Tìm x biết :
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
Tìm x biết: \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
Bài 4.1: Tìm x, biết
a) \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b) \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
c) \(\left|2\frac{1}{5}-x\right|+\left|x-\frac{1}{5}\right|+8\frac{1}{5}=1,2\)
d) \(2\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}=\left|2\frac{1}{5}-x\right|\)
Tìm x biết
a,\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|b,\left(x-7\right)^{x+1}+\left(x-7\right)^{x+11}=0\)
tìm x biết \(\left(x^2-1\right)\left(x^2-3\right)\left(x^2-5\right)\left(x^2-7\right)\le0\)
Tìm GTNN của các biểu thức :
a ) \(A=\left|x-1\right|+\left|x-2\right|+2016\)
b ) \(B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
