\(\frac{1}{3}+\frac{1}{6}=\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{499}{1000}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{499}{1000}\)

\(2\left(\frac{1}{2}-\frac{1}{x-1}\right)=\frac{499}{1000}\)

\(\frac{1}{2}-\frac{1}{x-1}=\frac{499}{2000}\)

\(\frac{2000\left(x-1\right)}{2\left(x-1\right)2000}-\frac{2.2000}{\left(x-1\right)2.2000}=\frac{499.2\left(x-1\right)}{2000.2\left(x-1\right)}\)

Khử mẫu 

\(2000x-2000-4000=998x-998\)

\(2000x-6000=998x-998\)

\(1002x-5002=0\)

\(1002x=5002\Leftrightarrow x=\frac{2501}{501}\)

Ta có: \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=1-\frac{2}{x+1}=\frac{2009}{2011}\)

\(\Rightarrow x=2010\).

Chúc em học tập tốt :)

ta có cái gì vậy chị huyền

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$

câu a+b dùng quy tắc chuyển vế

c, 3.(1/2-x)-5.(x-1/10)=-7/4

=>(3.1/2-3x)-(5x-5.1/10)=-7/4

=>3/2-3x-5x+1/2=-7/4

=>(3/2+1/2)-(3x+5x)=-7/4

=> 2-8x=-7/4

=>8x=15/4

=>x=15/4:8

=>x=15/32

a) 2.(1/4 - 3x) = 1/5 - 4x
=> 1/2 - 6x = 1/5 -4x
=> -6x + 4x = 1/5 - 1/2
=> -2x         = -3/10 = 3/20
b) 4.(1/3 - x) + 1/2 = 5/6 +x 
=> 4/3 - 4x + 1/2 = 5/6 +x
=> -4x - x = 5/6 - 4/3 - 1/2
=> -5x = -1
=> x= 1/5
c) 3. (1/2 - x) -5. ( x - 1/10) = -7/4
=> 3/2 - 3x - 5x + 1/2 = -7/4
=> -3x - 5x = -7/4 - 3/2 - 1/2 
=> -8x = -15/4
=> x = 15/32

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\left(\text{​Quy đồng nhé !}\right)\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{800}\)

=> x + 1 = 800 

<=> x = 799

1/3+1/6+1/10+...+1/x(x+1):2=399/400

2.[1/3.2+1/6.2+1/10.2+...+1/x(x+1)]=399/400

2.[1/6+1/12+1/20+...+1/x(x+1)]=399/400

2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=399/400

1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=399/800

1/2-1/x+1=399/800

1/x+1=1/800

=> x+1=800

=> x=799

a) (x+10)(2y-5) = 143

=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}

\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)

\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)

\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)

\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)

Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)

b) x+(x+1)+(x+2)+..+(x+30)=1240

=> (x+x+x+...+x)+(1+2+3+...+30)=1240

=> 31x+465=1240

31x = 1240-465

31x = 775

x = 775 : 31

x= 25

c) 1+2+3+...+x=210

\(\frac{\left(x-1\right)}{1}+1=x\)

=> \(\frac{\left(x+1\right).x}{2}=210\)

(x+1)x = 210:2

(x+1)x = 105

chắc ko có x thõa mãn

d) 2+4+6+...+2x=210

=> 2(1+2+3+...+x)=210

1+2+3+..+x= 210:2 = 105

\(\frac{\left(x-1\right)}{1}+1\) = x

\(\frac{\left(x+1\right).x}{2}=105\)

(x+1)x = 105:2

(x+1)x = 52,5

ko có x thõa mãn đề bài

a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}

 b, x+(x+1)+(x+2)+........+(x+30) = 1240

=> x+x+1+x+2+...+x+30=1240

=> 31x+(1+2+...+30) = 1240

=> 31x + 465 = 1240

=> 31x = 775

=> x = 25

c, 1+2+...+x=210

=> \(\frac{x\left(x+1\right)}{2}=210\)

=> x(x+1) = 420

Mà 420 = 20.21

=> x = 20

d, 2+4+...+2x = 210

=> 2(1+2+...+x) = 210

=> \(\frac{2x\left(x+1\right)}{2}=210\)

=> x(x + 1) = 210

Mà 210 = 14.15

=> x  = 14

e, 1+3+5+...+(2x-1) = 225

=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)

=> \(\frac{2x^2}{2}=225\)

=> x² = \(\left(\pm15\right)^2\)

=> x = 15 hoặc x = -15

Yêu cầu của đề bài là gì em? Với em để nhầm môn rồi, chuyển về đúng môn để được hỗ trợ tốt nhất hi.

ò chắc em nhầm môn tv á còn đề bài là viết thành số thập phân ý