Cho A = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+...+\dfrac{1}{2015.2016.2017}\)
So sánh A với \(\dfrac{1}{4}\)
Giải phương trình:
( \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}\)).x = (1.2+2.3+...+2006.2007)
\(L_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\)
\(L_2=1.2+2.3+...+2006.2007\)
\(3L_2=1.2.3+2.3.\left(4-1\right)+...+2006.2007.\left(2008-2005\right)\)
\(3L_2=1.2.3+2.3.4-1.2.3+...+2006.2007.2008-2005.2006.2007\)\(3L_2=2006.2007.2008\)
\(L_2=\dfrac{2006.2007.2008}{3}\)
\(pt\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\right).x=\dfrac{2006.2007.2008}{3}\)
Dễ dàng tìm được x nhé
Cho A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2014.2015.2016}\).So sánh A với \(\dfrac{1}{4}\)
Giúp mình nha!!
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2014.2015.2016}\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2015.2016}\right)\)
\(A=\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\)
\(=>A< \dfrac{1}{4}\)
Chúc bn học tốt
Ta có \(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3};\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{2}{2.3.4};\dfrac{2}{3.4.5};...;\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}=\dfrac{2}{2014.2015.2016}\)
2A= \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2014.2015.2016}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{5.6}-\dfrac{1}{6.7}+...+\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\)
A=\(\left(\dfrac{1}{2}:2\right)-\left(\dfrac{1}{2015.2016}:2\right)\)
A= \(\dfrac{1}{4}-\dfrac{1}{2015.2016.2}< \dfrac{1}{4}\)
Vậy A<\(\dfrac{1}{4}\)
Cho A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
Chứng minh A<2
Cho \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..............+\dfrac{1}{18.19.20}\) Chứng minh \(A< \dfrac{1}{4}\)
Help me!!!!!!!
\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\)
Theo công thức:
\(\dfrac{2m}{b.\left(b+m\right).\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+m.2\right)}\)Ta có:
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{18.19.20}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)2A=\(\dfrac{1}{1.2}-\dfrac{1}{19.20}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{19.20}\)
A=\(\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\)
A=\(\dfrac{1}{2}.\dfrac{19.20-2}{2.19.20}\)
A=\(\dfrac{19.20-2}{2.2.19.20}\) < \(\dfrac{19.20}{2.2.19.20}\) = \(\dfrac{1}{4}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
mik xin loi phan Ta có
\(\dfrac{2m}{b.\left(b+m\right)\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+2m\right)}\)Ta có blablabla
Tính A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{37.38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1482}\right)\)
\(A=\dfrac{1}{2}.\dfrac{370}{741}=\dfrac{185}{741}\)
dùng công thức \(\dfrac{2m}{a\left(a+m\right)\left(a+2m\right)}=\dfrac{1}{a\left(a+m\right)}-\dfrac{1}{\left(a+m\right)\left(a+2m\right)}\)để chứng tỏ rằng:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)
\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{19.20}< \dfrac{1}{1.2}\)
\(\Rightarrow2A< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{4}\) (đpcm)
dùng công thức \(\dfrac{2m}{a\left(a+m\right)\left(a+2m\right)}=\dfrac{1}{a\left(a+m\right)}-\dfrac{1}{\left(a+m\right)\left(a+2m\right)}\)để chứng tỏ rằng:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
1, Chứng minh
a) A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
b) B=\(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+....+\dfrac{36}{25.26.27}< 3\)
a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
tinh nhanh
A = \(\dfrac{1}{1.2.3}\)*\(\dfrac{1}{2.3.4}\)*................................*\(\dfrac{1}{18.19.20}\)