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|x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x
Do |x + 1| + |x + 2| + |x + 3| + |x + 4| \(\ge\)0
=> 5x \(\ge\)0
=> x + 1 + x + 2 + x + 3 + x + 4 = 5x
=> 4x + 10 = 5x
=> x = 10
P/s : Sai thì cậu thông cảm cho mình nha :P
=| x + 1+ x + 2 + x + 3 + x +4 | =5x
=|4x +( 1+ 2 + 3 + 4 )| =5x
=|4x + 10| =5x
=4x + 10 = 5x
=10 = 5x : 4x
=10 = x
=>x = 10
28373822839999999999999399393393933939939393393939393939393939933939393939393939393993939393939399393
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=1.\left(1-\frac{1}{101}\right)\)
\(B=\frac{100}{101}\)
\(C=\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{73.76}\)
\(C=4.\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
\(C=4.\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=\frac{4}{3}.\frac{9}{38}\)
\(C=\frac{6}{19}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\\ =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =1-\frac{1}{100}\\ =\frac{99}{100}\\ B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\\ =\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\\ =\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{4}{4.7}+\frac{4}{7.10}+....+\frac{4}{73.76}\\ =\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{73.76}\right)\\ =\frac{4}{3}.\left(\frac{3}{4}-\frac{3}{76}\right)\\ =\frac{18}{19}\)
Học tốt Nghe!!
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
=>\(3A=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{91.94}+\frac{3}{94.97}\)
=>\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{13}+...+\frac{1}{91}-\frac{1}{94}+\frac{1}{94}-\frac{1}{97}\)
=>\(3A=1-\frac{1}{97}\)
=>3A=\(\frac{96}{97}\)
=>A=\(\frac{32}{97}\)
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
1.
a) \(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)
b) Sửa đề: B = 1/2.5 + 1/5.8 + 1/8.11 + ...
\(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\\ B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\\ B=\frac{1}{6}-\frac{1}{294}\\ B=\frac{49}{294}-\frac{1}{294}=\frac{48}{294}=\frac{8}{49}\)
2.
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ \frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ 2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{2000}:2\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4000}\\ \frac{1}{2}-\frac{1999}{4000}=\frac{1}{n+1}\\ \frac{1}{n+1}=\frac{1}{4000}\\ \Rightarrow n+1=4000\\ \Rightarrow n=3999\)
Vậy n = 3999
41875 = 4. 10\(^4\) + 1.10\(^3\) + 8.10\(^2\) + 7.10 + 5.10\(^0\)