\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).5x+6}=\frac{2005}{2006}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\frac{1}{5x+6}=1-\frac{2005}{2006}\)

\(\frac{1}{5x+6}=\frac{1}{2006}\)

=>5x+6 = 2006

=>5x = 2000

=>x = 400

x = 400 nha bạn

                       Bài làm :

Ta có :

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)

\(\Leftrightarrow\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{5x+6-5x-1}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)

\(\Leftrightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Leftrightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Leftrightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)

\(\Leftrightarrow\frac{1}{5x+6}=\frac{1}{2006}\)

\(\Leftrightarrow1.2006=\left(5x+6\right).1\)

\(\Leftrightarrow2006=5x+6\)

\(\Leftrightarrow5x=2006-6\)

\(\Leftrightarrow5x=2000\)

\(\Leftrightarrow x=2000\div5=400\)

Vậy x=400

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

<=>/x/-4-2+/-x/=1/3/x/-5

<=>/x/-6+/-x/=1,3/x/-5

<=>-0,3/x/+/-x/=1

Th1 với x<0

=>/x/=-x ; /-x/=x

khi đó ta có:

-0,3.(-x) + x=1

<=>0,3x + x=1

<=>1,3x=1

<=>x=10/13

TH2 \(x\ge0\)

/x/=x;/-x/=-x

Khi đó ta có

-0,3x-x=1

<=>-1,3x=1

<=>x=-10/13

k nha

2.(x+5) - x² - 5x = 0

2(x+5) - x(x+5) = 0

(x+5)(2-x) = 0

=> x+5=0 hoặc 2-x=0

=> x=-5 hoặc x=2

\(\frac{x^5-1}{x^3-1}=\frac{ }{x+1}\)

Lời giải:

Nếu $x\geq 2$ thì:

$x-2-5(x-4)=20-5x$

$x-2=20-5x+5(x-4)=0$

$\Rightarrow x=2$ (tm) 

Nếu $x<2$ thì:

$2-x-5(x-4)=20-5x$

$2-x=20-5x+5(x-4)=0$

$\Rightarrow x=2$ (loại do $x<2$)

Vậy $x=2$