Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)
\(\Leftrightarrow x+3=103\)
\(\Leftrightarrow x\)\(=103-3\)
\(\Leftrightarrow x\)\(=100\)
Vậy x = 100
~~~~~~~Hok tốt~~~~~~~~
ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)
\(\Rightarrow x+3=103\)
\(\Rightarrow x=100\)
nhớ k nha
a) \(\frac{1,11+0,19-12,2}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
Đề đúng:
\(\frac{1,11+0,19-1,3.2^6}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
\(\frac{1,3-1,3.64}{0,39}-x=\left(\frac{3}{6}+\frac{2}{6}\right).\frac{1}{2}\)
\(\frac{1,3.\left(-63\right)}{0,39}-x=\frac{5}{6}.\frac{1}{2}\)
\(\frac{-81,9}{0,39}-x=\frac{5}{12}\)
\(\left(-210\right)-x=\frac{5}{12}\)
\(x=\left(-210\right)-\frac{5}{12}\)
\(x=\frac{-2520}{12}-\frac{5}{12}\)
\(x=\frac{-2525}{12}\)
b) \(x:\left(3\frac{1}{2}-5\frac{1}{6}\right)=4\frac{1}{5}-6\frac{2}{3}\)
\(x:\left(\frac{7}{2}-\frac{31}{6}\right)=\frac{21}{5}-\frac{20}{3}\)
\(x:\left(\frac{21}{6}-\frac{31}{6}\right)=\frac{63}{15}-\frac{100}{15}\)
\(x:\frac{-10}{6}=\frac{-37}{15}\)
\(x.\frac{6}{-10}=\frac{-37}{5}\)
\(x.\frac{-6}{10}=\frac{-37}{5}\)
\(x.\frac{-3}{5}=\frac{-37}{5}\)
\(x=\frac{-37}{5}:\frac{-3}{5}\)
\(x=\frac{-37}{5}.\frac{5}{-3}\)
\(x=\frac{-37}{-3}\)
\(x=\frac{37}{3}\)
Sai đề bài
Lẽ ra câu a chỉ chia 2 thì chia 22
Sai đề bài mina-san
a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
=> (2,85 - x - 0,5) : 0,25 = 60
=> (2,85 - 0,5) - x = 60 . 0,25
=> 2,35 - x = 15
=> x = 2,35 - 15
=> x = -12,65
Vậy x = -12,65
b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)
\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)
\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)
\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)
\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)
\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)
\(\Rightarrow x=\frac{26}{6}\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}\)
c) \(35\left(2\frac{1}{5}-x\right)=32\)
\(\Rightarrow2\frac{1}{5}-x=32\div35\)
\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)
\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)
\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)
\(\Rightarrow x=\frac{56}{405}\)
Vậy \(x=\frac{56}{405}\)
e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)
\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow4.\frac{12}{11}=11-5\div x\)
\(\Rightarrow11-5\div x=\frac{48}{11}\)
\(\Rightarrow5\div x=11-\frac{48}{11}\)
\(\Rightarrow5\div x=\frac{73}{11}\)
\(\Rightarrow x=5\div\frac{73}{11}\)
\(\Rightarrow x=\frac{55}{73}\)
Vậy \(x=\frac{55}{73}\)
a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
(2,85 - x - 0,5) : 0,25 = 60
(2,85 - x - 0,5) = 60 x 0,25
(2,85 - x - 0,5) = 15
2,35 - x = 15
x = 2,35 - 15
x = -12,65
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\)\(\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Rightarrow x+1=2001\)
\(\Rightarrow x=2001-1=2000\)
Vậy \(x=2000.\)
Chỗ \(x\) phải là \(\frac{2}{x\left(x+1\right)}\) chứ bạn :)
Ta có :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\) ( nhân hai vế cho \(\frac{1}{2}\) )
\(\Leftrightarrow\)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2001}\)
\(\Leftrightarrow\)\(x-1=2001\)
\(\Leftrightarrow\)\(x=2001+1\)
\(\Leftrightarrow\)\(x=2002\)
Vậy \(x=2002\)
Chúc bạn học tốt ~
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
Mình ghi nhầm đó, lẽ ra phải thế này cơ:\(\left(30:7\frac{1}{2}+0,5x3-1,5\right)x\left(4\frac{1}{2}-\frac{9}{2}\right):\left(14,5x100\right)\)
\((30:7\frac{1}{2}\)\(+0,5x3\)\(-1,5)x(4\frac{1}{2}\)\(-\frac{9}{2}\)\():(14,5x100)\)
\(=\)\((30:7\frac{1}{2}\)\(+0,5x3\) \(-1,5)x(\frac{9}{2}\)\(-\frac{9}{2}\)\()\)\(:(14,5x100)\)
\(=30:7\frac{1}{2}\)\(+0,5x3-1,5)x0x(14,5x100)\)
\(=0\)
Chúc bạn học tốt!
Bài 2: ta có tích riêng thứ nhất là .....5, thứ hai cũng là ....5 -> chữ số tận cùng là: ....5 - ....5 = ...0
Bài 3: Gọi số có hai chữ số đó là ab (a,b =<9)
...........................__..... _
Theo đề bài ta có: ab = 9b
=> b = (2; 3; 4; 5; 6; 7; 8; 9)
..........................................
=> Tương ứng với b ta có ab = (18; 27; 36; 45; 54; 63; 72; 81)
Nhận xét: Chỉ có 45 = 9.5
Vậy số đó là 45
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}=1\frac{9}{11}\)
=>\(\left\{1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}\right\}\times\frac{1}{2}=1\frac{9}{11}\times\frac{1}{2}\)
=>\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)
=>\(1-\frac{1}{x+1}=\frac{10}{11}\)
=> \(\frac{1}{x+1}=1-\frac{10}{11}\)
=> \(\frac{1}{x+1}=\frac{1}{11}\)
=> x + 1 = 11
=> x = 10
Nhấn đúng cho mk nha^^