N
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
11 tháng 9 2019
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
11 tháng 9 2019
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)
29 tháng 7 2016
\(a,x-\frac{5}{6}:1\frac{1}{6}=0,125\)
\(x-\frac{5}{6}:\frac{7}{6}=\frac{1}{8}\)
\(x-\frac{5}{7}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{7}\) \(x=\frac{47}{56}\)
\(b,\left(1-\frac{2}{10}+x+\frac{1}{5}\right):\left(1\frac{1}{3}-\frac{2}{3}+3\frac{1}{3}\right)-1=1\frac{1}{2}\)
\(\left(1-\frac{1}{5}+x+\frac{1}{5}\right):\left(\frac{4}{3}-\frac{2}{3}+\frac{10}{3}\right)-1=\frac{3}{2}\)
\(\left(\frac{4}{5}+x+\frac{1}{5}\right):4=\frac{3}{2}+1\)
\(\left(1+x\right):4=\frac{5}{2}\)
\(1+x=\frac{5}{2}.4\)
\(1+x=10\)
\(x=10-1\)
\(x=9\)
21 tháng 7 2020
Tìm X
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)
\(=\frac{1}{5}\)
\(\Rightarrow x-100=\frac{1}{5}\)
\(x=\frac{1}{5}+100\)
\(x=\frac{1}{5}+\frac{500}{5}\)
\(x=\frac{501}{5}\)
TN
28 tháng 7 2018
\(2\frac{3}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{8}-\frac{3}{14}+\frac{1}{3}\)
\(=\frac{251}{168}\)
T
28 tháng 7 2018
Bài 1 : a, thực hiện phép tính :
\(2\frac{3}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{1}{2}×\left(\frac{11}{4}-\frac{3}{7}\right)+\frac{1}{3}\)
\(=\frac{1}{2}×\frac{65}{28}+\frac{1}{3}\)
\(=\frac{65}{56}+\frac{1}{3}\)
\(=\frac{251}{168}\)
b , Tìm x biết :
a, 435- ( x + 16 ) = 425 : 17
435 - ( x + 16 ) = 25
x + 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
Vậy x = 394
b, ( x + 3/4 ) × 7/4 = 5 - 7/6
( x + 3/4 ) × 7/4 = 23/6
x + 3/4 = 23/6 : 7/4
x + 3/4 = 23/6 × 4/7
x + 3/4 = 46/21
x = 46/21 - 3/4
x = 121/84
Vậy x = 121/84
TS
1
NN
0
K
18 tháng 11 2018
\(a,\frac{2}{3}=\frac{x}{54}\)
\(\Rightarrow2.54=3x\)
\(\Rightarrow3x=108\)
\(\Rightarrow x=108:3=36\)
\(b,\frac{10}{x}=\frac{15}{6}\)
\(\Rightarrow10.6=15x\)
\(\Rightarrow15x=60\)
\(\Rightarrow x=60:15=4\)
\(c,\frac{2}{3}< \frac{x}{6}< 1\)
\(\Rightarrow\frac{4}{6}< \frac{x}{6}< \frac{6}{6}\)
\(\Rightarrow4< x< 6\)
\(\Rightarrow x=5\)
\(d,1< \frac{6}{x}< 2\)
\(\Rightarrow\frac{6}{6}< \frac{6}{x}< \frac{6}{3}\)
\(\Rightarrow6< x< 3\)
\(\Rightarrow x=5;4\)
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