a, (x+2)+(x+4)+(x+6)+...+(x+100)=6000

(x+x+x+...+x)+(2+4+6+...+100)=6000

50.x+2550=6000

50.x=6000-2550

50.x=3450

x=3450:50

x=69

b, 1+2+3+4+...+x=15

10+...+x=15

x=15-10

x=5

Nho **** cho minh nha

Ta có: (x+x+x+...+x) + (2+4+6+...+100) = 6000

Ta thấy vế phải có: (100-2):2+1=50(số hạng)

Tổng của vế phải: [(2+100)*50]:2=2550

\(\Rightarrow\)có 50 số x

\(\Rightarrow\)50*x + 2550 = 6000

\(\Rightarrow\)50*x=6000-2550

\(\Rightarrow\)50*x=3450

\(\Rightarrow\)x=3450:50

\(\Rightarrow\)x=69

   Vậy x=69

Mình đúng nè, nhớ k nha

njghjhjkhjk

a. nhân cả hai vế của đẳng thức với 1/ 10 ta có

x/10 - (2/11.13 +2/13.15+...+2/53.55)=3/11 . 1/10

x/10 - (1/11-1/13+1/13-1/15 +...+1/53-1/55) =3/110

x/10 - (1/11 - 1/55) =3/110

x/10 -4/55 = 3/110

x/10=3/110 + 4/55

x. 1/10 =1/10

x= 1/10 : 1/10 =1

b) bạn nhân cả hai vế của đẳng thức với 1/2 rồi làm tương tự

a. nhân cả hai vế của đẳng thức với \(\frac{1}{10}\). Ta có:

\(\frac{x}{10}-\left(\frac{2}{11.13}+\frac{2}{13.15}+...\frac{2}{53.55}\right)=\frac{3}{11}.\frac{1}{10}\)

\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{110}\)

\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{110}\)

\(\frac{x}{10}-\frac{-4}{55}=\frac{3}{110}\)

\(\frac{x}{10}=\frac{3}{110}+\frac{4}{55}\)

\(x.\frac{1}{10}=\frac{1}{10}\)

\(x=\frac{1}{10}:\frac{1}{10}=1\)

b. cũng thế bạn nhân hai vế của đẳng thức với \(\frac{1}{2}\) rồi làm tương tự.

x - \(10\left(\frac{2}{11.13}+\frac{2}{13.15}+..+\frac{2}{53.55}\right)=22\)

\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+..+\frac{1}{53}-\frac{1}{55}\right)=22\)

\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=22\)

\(x-10\cdot\frac{4}{55}=22\)

\(x-\frac{40}{55}=22\)

\(x=22+\frac{40}{55}=\frac{250}{11}\)

\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-......-\frac{20}{53\cdot55}=\frac{3}{11}\)

\(\Leftrightarrow x-10\left(\frac{2}{11\cdot13}-\frac{2}{13\cdot15}-\frac{2}{15\cdot17}-.....-\frac{2}{53\cdot55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+....+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x=1\)

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Rightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\left[10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{20}{53.55}\right)\right]=\frac{3}{11}\)

\(\Rightarrow x-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Rightarrow x-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Rightarrow x-\left[10.\frac{4}{55}\right]=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}\)

\(\Rightarrow x=1\)

Vậy x = 1

_Chúc bạn học tốt_

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\Rightarrow x=\frac{3}{11}+\frac{8}{11}=\frac{11}{11}=1\)

\(x-\left(\frac{20}{11\cdot13}+\frac{20}{13\cdot15}+...+\frac{20}{53\cdot55}\right)=\frac{3}{11}\)

Đặt \(x-A=\frac{3}{11}\)

\(\frac{A}{10}=\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\)

\(\frac{A}{10}=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)

\(\frac{A}{10}=\frac{1}{11}-\frac{1}{55}\)

\(\frac{A}{10}=\frac{4}{55}\)

\(A=\frac{8}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}\)

\(\Rightarrow x=1\)

bài 1:

\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{231}{4}\cdot\frac{4}{21}=11\)

\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)

= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)

= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)

= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)

= \(\frac{31}{34}:2\)

= \(\frac{31}{68}\)

đề đúng

\(\frac{7}{4}\cdot\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

x-10( 2/11.13+2/13.15+...+2/53.55)=3/11

x-10(1/11-1/55)=3/11

x-10.4/55=3/11

x-40/55=3/11

x=3/11+40/55

x= 1

x-10.(2/11.13+2/13.15+....+2/53.55)=3/11

x-10.(1/11-1/13+...+1/53-1/55)=3/11

x-10.(1/11-1/55)=3/11

x-10.4/55=3/11

x-8/11=3/11

x=1

Nhớ k cho mình nhé