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đặt VT là A ta đc:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(A=\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$
= $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$
= $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$
= $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$
= $\frac{41}{3.39}$
D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)
= \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)
= \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)
= \(\frac{41}{117}\)
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19
1/3.(1-1/x+3)=6/19
1-1/x+3=6/19:1/3
1-1/x+3=18/19
1/x+3=1-18/19
1/x+3=1/19
=> x+3=19
=>x=19-3
x=16
Đặt biểu thức là A, ta có:
3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)
3A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
3A=1-\(\frac{1}{x+3}\)
A=\(\frac{1}{3}-\frac{3}{x+3}\)
=>\(\frac{1}{3}-\frac{3}{x+3}\) =\(\frac{6}{19}\) =>x=168
a) Để A nguyên => 5 chia hết cho n - 2
n - 2 thuộc U(5) = {-5 ; -1 ; 1 ; 5}
n - 2 = -5 => n = -3
n - 2 = -1 => n = 1
n - 2 = 1 => n = 3
n - 2 = 5 => n = 7
Vậy n thuộc {-3 ; 1 ; 3 ; 7}
b) \(\frac{y}{3}-\frac{1}{x}=\frac{1}{3}\Leftrightarrow\frac{y}{3}-\frac{1}{3}=\frac{1}{x}\)
\(\frac{y-1}{3}=\frac{1}{x}\) <=> (y-1).x = 3
(y-1).x = 1.3 = (-1).(-3)
TH1: y - 1 = 1 => y = 2
=> x = 3
TH2: y - 1 = 3 => y = 4
=> x = 1
TH3: y - 1 = -1 => y = 0
=> x = -3
TH4: y - 1 = -3 => y = -2
=> x = -1
Vậy (x ; y) là (2 ; 3) ; (4 ; 1) ; (0 ; -3) ; (-2 ; -1)
a) Để A là 1 số nguyên thì n-2 \(\in\) Ư(5)={-1;-5;1;5}
Nếu n-2=-1 thì n=1
Nếu n-2=-5 thì n=-3
Nếu n-2=1 thì n=3
Nếu n-2=5 thì n=7
=>n \(\in\) {-3;1;3;7}
b) câu b này mik ko biết làm 
Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)
\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)
\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> 2015 + x = 0
=> x = -2015
A=1-1/4+1/4-1/7+1/7-1/10+....+1/n-1/(n+3)
A=1-1/(n+3)
vì 1/(n+3)lớn hơn 0 nên 1-1/(n+3)<1
=>A<1
(x-1+3)/9=1/y+2
(x+2)/9=1/(y+2)
tích chéo:x.y+2x+2y=5
phân phối ra rồi tìm ước của 5 sau đó lập bảng là ra