Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
Câu hỏi của Cristiano Ronaldo - Toán lớp 7 - Học toán với OnlineMath
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
(1/4-1).(1/5-1).....(1/2000-1).(1/2001-1)
=(-3/4).(-4/5).(-5/6).....(-1999/2000).(-2000/2001)
=-3.-4.-5....-1999.-2000/4.5.6...2000.2001
=-3/2001
\(-\frac{1}{2003.2002}-\frac{1}{2002.2001}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-\left(1-\frac{1}{2003}\right)\)
\(=\frac{-2002}{2003}\)
\(\frac{-1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2000.2001}+\frac{1}{2001.2002}+\frac{1}{2002.2003}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-\left(1-\frac{1}{2003}\right)\)
\(=-\frac{2002}{2003}\)