\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\) giúp em tìm công thức với ạ
Những câu hỏi liên quan
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+2\right)}=\frac{n}{6n+4}\)
b.\(\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\cdot\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
Đúng 0
Bình luận (0)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
Đúng 0
Bình luận (0)
Tìm x biết: \(\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{65\cdot68}\right)x-\frac{7}{34}=\frac{19}{68}\)
\(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{65.68}\right)x=\frac{19}{68}+\frac{7}{34}\)
\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\left(\frac{1}{2}-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\frac{33}{68}x=\frac{33}{68}\)
\(x=\frac{33}{68}:\frac{33}{68}=1\)
Đúng 0
Bình luận (0)
Tìm x biết :\(\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{65\cdot68}\right)x-\frac{7}{34}=\frac{19}{68}\)
Tìm x biết: \(\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{65\cdot68}\right)x-\frac{7}{34}=\frac{19}{68}\)
(1/2.5+1/5.8+......+1/65.68)x=19/68+7/34=33/68
(1/2.5+1/5.8+.......+1/65.68).x.3=33/68.3=99/68
(3/2.5+3/5.8=.......+3/65.68).x=99/68
(1/2-1/5+1/5-1/8+....-+1/65-1/68).x=99/68
(1/2-1/68).x=99/68
33/68.x=99/68
=>x=3
k cho mình nha
Đúng 0
Bình luận (0)
Tìm số nguyên âm a biết:
\(a^2-\left(\frac{3}{5}\right)^2=\frac{1}{1\cdot2}+\frac{1}{2\cdot7}+\frac{1}{7\cdot5}+\frac{1}{5\cdot13}+\frac{1}{13\cdot8}+\frac{1}{8\cdot19}+\frac{1}{19\cdot11}+\frac{1}{11\cdot25}\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Đúng 0
Bình luận (0)
Tìm số nguyên âm a biết:
\(a^2-\left(\frac{3}{5}\right)^2=\frac{1}{1\cdot2}+\frac{1}{2\cdot7}+\frac{1}{7\cdot5}+\frac{1}{5\cdot13}+\frac{1}{13\cdot8}+\frac{1}{8\cdot19}+\frac{1}{19\cdot11}+\frac{1}{11\cdot25}\)
Tìm số nguyên dương a thoả mãn:
\(a^2-\left(\frac{3}{5}\right)^2=\frac{1}{1\cdot2}+\frac{1}{2\cdot7}+\frac{1}{7\cdot5}+\frac{1}{5\cdot13}+\frac{1}{13\cdot8}+\frac{1}{8\cdot19}+\frac{1}{19\cdot11}+\frac{1}{11\cdot25}\)
Ai làm nhanh nhất thì mình tick cho
\(a^2\)- (\(\frac{3}{5}^2\)) = \(\frac{1}{1}\)-\(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{19}\)+\(\frac{1}{19}\)-\(\frac{1}{11}+\frac{1}{11}\)\(-\frac{1}{25}\)
= 1\(-\frac{1}{25}\)
= \(\frac{24}{25}\)
chúc bạn học tốt
Đúng 0
Bình luận (0)
1. Chứng minh rằng với \(\forall N\ne0̸\) ta đều có :
a, \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{\left(3n-1\right)\cdot\left(3n+1\right)}=\dfrac{n}{6n+4}\).
2. Tìm GTLN hoặc GTNN của biểu thức \(A=\dfrac{\left|2-x\right|-3}{\left|2-x\right|+11}\).
\(\frac{1}{\left(3n+2\right)\left(3n+5\right)}=\frac{3n+5-3n-2}{\left(3n+2\right)\left(3n+5\right)}=\frac{3}{\left(3n+2\right)\left(3n+5\right)}giainhuthedungko\)sai sử giúp nhé thank
Chắc có lẽ bạn định làm như này:
\(\frac{1}{\left(3n+2\right)\left(3n+5\right)}=\frac{3}{3\left(3n+2\right)\left(3n+5\right)}=\frac{\left(3n+5\right)-\left(3n+2\right)}{3\left(3n+2\right)\left(3n+5\right)}=\frac{1}{3}\left[\frac{1}{3n+2}-\frac{1}{3n+5}\right]\)