Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^

duyệt đi

Đơn giản -_-

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)

(=)\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+..+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)

\(\Rightarrow\)x=305

a) \(\dfrac{5}{8}-\left(x+\dfrac{3}{4}\right)=\dfrac{7}{6}\)

     \(-x-\dfrac{3}{4}=\dfrac{13}{24}\)

    \(-x=\dfrac{31}{24}\)

\(\Rightarrow x=\dfrac{-31}{24}\)

b) \(\left(x-\dfrac{3}{4}\right)+\dfrac{3}{2}=\dfrac{11}{3}\)

    \(x-\dfrac{3}{4}=\dfrac{13}{6}\)

    \(x=\dfrac{35}{12}\)

c) \(\left(x+\dfrac{1}{2}\right)-\dfrac{5}{8}=2\)

    \(x+\dfrac{1}{2}=\dfrac{21}{8}\)

    \(x=\dfrac{17}{8}\)

d) \(\dfrac{11}{2}-\left(\dfrac{4}{5}+x\right)=2-\dfrac{1}{3}\)

    \(\dfrac{11}{2}-\left(\dfrac{4}{5}+x\right)=\dfrac{5}{3}\)

    \(-\dfrac{4}{5}-x=\dfrac{-23}{6}\)

    \(-x=\dfrac{-91}{30}\)

\(\Rightarrow x=\dfrac{91}{30}\) 

`5/8-(x+3/4)=7/6`

`=> x+3/4 = 5/8 -7/6`

`=> x+3/4 = -13/24`

`=>x= -13/24 -3/4`

`=>x= -13/24 - 18/24`

`=>x= -31/24`

`-----------`

`(x-3/4)+3/2+11/3`

`=11/3` hả bn?

`----------`

`(x+1/2)-5/8=2`

`=>x+1/2=2+5/8`

`=>x+1/2= 16/8+5/8`

`=> x+1/2=21/8`

`=>x=21/8-1/2`

`=>x= 21/8 - 4/8`

`=>x= 17/8`

`---------`

`11/2-(4/5+x)=2-1/3`

`=> 11/2-(4/5+x)=6/3-1/3`

`=> 11/2-(4/5+x)=5/3`

`=>4/5+x=11/2-5/3`

`=>4/5+x=23/6`

`=>x=23/6-4/5`

`=>x= 91/30`

`(x-3/4) + 3/2 =11/3`

`=>x-3/4=11/3-3/2`

`=>x-3/4=13/6`

`=>x= 13/6 + 3/4`

`=>x= 35/12`

\(2\left(x-5\right)+3\left(2-3x\right)=5x+7\)

\(\Leftrightarrow2x-10+6-9x=5x+7\)

\(\Leftrightarrow\left(2x-9x\right)+\left(6-10\right)=5x+7\)

\(\Leftrightarrow-7x-4=5x+7\)

\(\Leftrightarrow-7x-5x=4+7\)

\(\Leftrightarrow-12x=11\)

\(\Leftrightarrow x=\frac{-11}{12}\)

\(3x-5\left(x-2\right)+7=4x-12\)

\(\Leftrightarrow3x-5x-10+7=4x-12\)

\(\Leftrightarrow\left(3x-5x\right)-\left(10-7\right)=4x-12\)

\(\Leftrightarrow-2x-3=4x-12\)

\(\Leftrightarrow-2x-4x=3-12\)

\(\Leftrightarrow-6x=-15\)

\(\Leftrightarrow x=\frac{-15}{-6}=\frac{5}{2}\)

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}=\frac{101}{1540}.3\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5\left(x+3\right)}=\frac{1}{5}-\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5x+15}=\frac{308}{1540}-\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5x+15}=\frac{5}{1540}\)

\(\Leftrightarrow5x+15=1540\)

\(\Leftrightarrow5x=1540-15\)

\(\Leftrightarrow5x=1525\)

\(\Leftrightarrow x=1525\div5\)

\(\Leftrightarrow x=305\)

\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+........+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{101}{1540}\)

3(.\(\frac{1}{5.8}+\frac{1}{8.11}\)+\(\frac{1}{11.14}+.......+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}.3=\frac{303}{1540}\)

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

=>\(x+3=308\)

\(x=308-3=305\)

Vậy \(x=305\)

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1}{308}\)

=> x + 3 = 308

     x = 308 - 5

     x = 303

\(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{5}+\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\left(\frac{1}{5}-\frac{1}{x+3}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{x}-\frac{1}{x}\right)=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

=>x+3=308

x=308-3

x=305

Vậy x=305

a, (x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75

 5x + (2 + 5 + 8 + 11 + 14) = 75

5x + 40 = 75

5x = 35

x = 7

Các câu dưới tương tự.

a)(x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75

x+x+x+x+x+2+5+8+11+14=75

5xx+40=75

5xx=75-40

5xx=35

x=35:5

x=7

b)(x+1)+(x+3)+(x+5)+....+(x+101)=5400

Nhận thấy dãy trên có (101-1):2+1=51(số) nên có 51 x

Tách như câu a

51xx+(1+3+5+...+101)=5400

51xx+[(101+1)x51:2]=5400

51xx+2601=5400

51xx=5400-2601

51xx=2799

x=2799:51

x=933/17

c)(x+1)+(x+4)+(x+7)+....+(x+94)=1840

Phần c này cũng giống phần b,em dựa vào mà làm nhé,giống,khác mỗi số thôi

Chúc em họ tốt^^