Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)
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
x-(20/11*13+20/13*15+20/15*17+...+20/553*55)=3/7
(a-b)(a-b)+(b-c)(b-c)+(c-a)(c-a)=(a+b-2c)(a+b-2c)+(b+c-2a)(b+c-2a)+(c+a-2b)(c+a-2b)
Cm:a=b=c