a: x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b: 2x(x+3)=0

=>x(x+3)=0

=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

c: \(\left(6-x\right)\left(x+10\right)=0\)

=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)

d: \(\left(5x+20\right)\left(x^2+1\right)=0\)

=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)

=>5x=-20

=>x=-4

Ta có:

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

\(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)

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

\(x=-2012\)

;Ko tồn tại nghiệm số thực OK

a.|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=4 hoăc x=-2

vậy...

b.2x+17=15

=>2x=15-17

=>2x=-2

=>x=-1

vậy...

c.tự giải

d.4x-15=-75-x

=>4x+x=-75+15

=>5x=-60

=>x=-12

vậy...

2.2+(-3)+4+...+(-2011)+2012

=(-3+2)+...+(-2011+2010)+2012 (có 2010 cặp)

=-1.2010+2012

=-2010+2012=2

Ai mún kb vs mink ko

2.

Tổng A có số số hạng  là:2012-2+1=2011(số)

A=[2+(-3)]+....+[2010+(-2011)]+2012

A=(-1)+..+(-1)+2012

Vậy sẽ có:1005 cặp số

A=[(-1)*1005]+2012

A=(-1005)+2012

A=1007

Nhớ k cho mình nhé

a) x + 9 = 2 . 17

   x  + 9 = 34

   x        = 34 - 9

   x        = 25

Vậy x = ...

b) x - 17 = ( - 11 ) . ( - 5 )

    x -17  = 55

    x        = 55 + 17 

    x        =  72

Vậy x =...

d) 2011 + 2010 +2009 + .......... + x = 2011

d) 2011 + (2010 +2009 + .......... + x) = 2011

2010 + 2009 +...+ x= 2011 -2011

2010 + 2009 +...+ x  = 0

=> Từ 2010 -> x có : x số

\(\rightarrow\frac{(x+2010).x}{2}=0\)

\(\Rightarrow x+2010.x=0.2\)

\(\Rightarrow x+2010.x=0\)

\(\Rightarrow x+2010=0\)

\(\Rightarrow x=0-2010\)

\(\Rightarrow x=-2010\)

Vậy x =...

`Answer:`

\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)

\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Rightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

x+5/2009 + x+4/2010 = x+3/2011 + x+2/2012

=> 1 + x+5/2009 + 1 + x+4/2000 = 1 + x+3/2011 + 1 + x+2/2012

=> x+2014/2009 + x+2014/2000 = x+2004/2011 + x+2014/2012

=> x+2014/2009 + x+2014/2000 - x+2014/2011 - x+2014/2012 = 0

=> (x+2014).(1/2009 + 1/2010 - 1/2011 - 1/2012) = 0

Do 1/2009 > 1/2011; 1/2010 > 1/2012

=> 1/2009 + 1/2010 - 1/2011 - 1/2012 khác 0

=> x + 2014 = 0

=> x = -2014

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)

=>x-2014=0

hay x=2014