\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)  

\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

k cho minh

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)

\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)

Tính ra nhé !

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=x+\frac{1}{6}\)

\(\Rightarrow4x+\frac{77}{60}=x+\frac{1}{6}\)

\(\Rightarrow3x=\frac{1}{6}-\frac{77}{60}\)

\(\Rightarrow3x=-\frac{67}{60}\)

\(\Rightarrow x=-\frac{67}{60}\div3=\frac{-67}{60.3}=-\frac{67}{180}\)

Vậy x = .........

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)

<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)

<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)

<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

<=> x - 2012 = 0

<=> x = 2012

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2013\right)=4+1007\cdot2013\)

\(\Leftrightarrow2014x+2027091=2027095\)

\(\Leftrightarrow2014x=4\)

hay \(x=\dfrac{2}{1007}\)

số số hạng: ( 2500 - 2 ) : 2 +1  = 1250 số

tổng: ( 2500 +2  ) . 1250 : 2 = 1563750

Vì: 1250 . 1251 = 1563750

=> x = 1250 

dat \(\frac{x}{5}=\frac{y}{4}=k\)-> x=5k va y=4.k

thay x=5k va y=4k vao x²-y²=1 ta duoc

(5k)²-(4k)²=1

25k²-16k²=1

9k²=1

k²=\(\frac{1}{9}=\left(\frac{1}{3}\right)^2\)

-> k=1/3 hay k=-1/3

voi K=1/3--> x=5.1/3=5/3 va y=4.1/3=4/3

voi K=-1/3->x=5.-1/3=-5/3 va y=4.-1/3=-4/3

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=4+1007\cdot2003\)

\(\Leftrightarrow2004x+\dfrac{2003\cdot2004}{2}=4+1007\cdot2003\)

\(\Leftrightarrow2004x=10019\)

hay \(x=\dfrac{10019}{2004}\)

\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Leftrightarrow x^3-25x-x^3-8=42\)

\(\Leftrightarrow-25x-8=42\)

\(\Leftrightarrow-25x=42+8\)

\(\Leftrightarrow-25x=50\)

\(\Leftrightarrow x=-\dfrac{50}{25}=-2\)

nhầm rồi

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