\(\frac{x}{2}\cdot3+x\cdot4=23,1\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+..+\frac{1}{8\cdot9\cdot10}\cdot x=\frac{22}{45}\)thì x=
B cho mình hỏi hình như thiếu dấu ngoặc
tìm x biết : (\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{8\cdot9\cdot10}\)) x \(x=\frac{22}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\) vậy
\(\frac{11}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}\div\frac{11}{45}=2\)
vậy suy ra x =2
mình chắc chắn 100% luôn đó, cái này ở trong violympic toán 7 vòng 12 phải ko
Tìm x biết: \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\cdot x=\frac{22}{45}\)Trình bày cách tính
X=2 nhé bạn.....đúng đó, vòng 12 mk 300 mà cx gặp câu này!!! Tick nha
giải pt
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.......+\frac{1}{2005.2006.2007}\right)x=\left(1\cdot2+2\cdot3+.....+2006.2007\right)\)
Tìm x biết: \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{22}{45}\).
Có công thức cho mình thì càng tốt!
Trước hết ta thực hiện biểu thức trong ngoặc:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)
\(=\frac{1}{2}.\frac{22}{45}\) \(=\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}\) \(.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)
\(\Rightarrow x=2\)
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
Tìm số tự nhiên x biết:
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+............+\frac{2}{x+\left(x+1\right)}=\frac{2008}{2010}\)
\(\frac{2}{2.3}\)+ \(\frac{2}{3.4}\)+ \(\frac{2}{4.5}\)+........+ \(\frac{2}{x+\left(x+1\right)}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+..........+ \(\frac{1}{x+\left(x+1\right)}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+.........+ \(\frac{1}{x}\)- \(\frac{1}{x+1}\)= \(\frac{2008}{2010}\)
= 2 . ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\)
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\): 2
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{2008}{2010}\). \(\frac{1}{2}\)
= ( \(\frac{1}{2}\)- \(\frac{1}{x+1}\)) = \(\frac{502}{1005}\)
= \(\frac{1}{x+1}\)= \(\frac{1}{2}\)- \(\frac{502}{1005}\)
= \(\frac{1}{x+1}\)= \(\frac{1}{2010}\)
\(\Rightarrow\)\(x+1\)= 2010
\(\Leftrightarrow\) \(x\) = 2010 - 1
\(\Rightarrow\) \(x\)= 2009
Vậy \(x\)= 2009
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x\left(x+1\right)}=\frac{2008}{2010}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1004}{1005}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
\(=>x+1=2010\)
\(=>x=2009\)
Vậy \(x=2009\)
Tìm số tự nhiên x, biết: \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
Gọi A=(1/1.2.3+ 1/2.3.4 +...+ 1/8.9.10) .x=23/45
2A=3-1/1.2.3+ 4–2/2.3.4+ 5–4/3.4.5+ ... + 10–8/8.9.10
2A=1/2 —1/2.3+ 1/2.3 — 1/3.4+ 1/3.4– 1/4.5 +...+1/8.9–1/9.10=1/2–1/9.10=44/90
A=44/90 : 2=22/90
x=23/45:A= 23/45 : 22/90=23/11= 2 1/1( hỗn số)