Tính tổng sau bằng cách hợp lí : A = \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{17\cdot20}\)
S= \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+......+\frac{1}{97\cdot100}\)
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{97.100}\)
\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{97.100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\frac{49}{100}=\frac{49}{300}\)
Ta có: \(S=\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{97.100}.\)
\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{97.100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow S=\frac{49}{100}:3=\frac{49}{300}\)
Vậy \(S=\frac{49}{300}\)
CHÚC BẠN HỌC TỐT
\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\)
\(S=3\cdot\frac{1}{3}\cdot\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\right)\)
\(S=\frac{1}{3}\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{97\cdot100}\right)\)
\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\cdot\left(\frac{50}{100}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\cdot\frac{49}{100}\)
\(S=\frac{49}{300}\)
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)
\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\) giúp em tìm công thức với ạ
\(3S=3\left(\frac{1}{2.5}+....+\frac{1}{\left(3n+1\right)\left(3n+2\right)}\right)\)
Đến đây thì bạn làm như dạng đơn giản nhé
Tìm x biết: \(\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{65\cdot68}\right)x-\frac{7}{34}=\frac{19}{68}\)
\(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{65.68}\right)x=\frac{19}{68}+\frac{7}{34}\)
\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\left(\frac{1}{2}-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\frac{33}{68}x=\frac{33}{68}\)
\(x=\frac{33}{68}:\frac{33}{68}=1\)
Cho S = \(\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+....+\frac{9}{29\cdot32}\). CM : S>1
Đề là cm S>1 nha bạn!
\(S=\frac{9}{2.5}+\frac{9}{5.8}+...+\frac{9}{29.32}\)
\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=3.\frac{15}{32}\)
\(=\frac{45}{32}>1\)
\(\Leftrightarrow S>1\)
\(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}\)
Cách 1 : Vì hiệu hai thừa số đều là 3 = 5 - 2 = 8 - 5 = ... = 32 - 29 nên phân tích tử 9 = 3 . 3
Ta có : \(S=3\left[\frac{3}{2\cdot5}+\frac{3}{7\cdot9}+...+\frac{3}{29\cdot32}\right]=3\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{32}\right]\)
\(=3\left[\frac{1}{2}-\frac{1}{32}\right]=3\left[\frac{16}{32}-\frac{1}{32}\right]=3\cdot\frac{15}{32}=\frac{45}{32}\)
Mà \(\frac{45}{32}>1\)=> S không thể bé hơn 1
Cách 2 : Nhận xét : \(\frac{9}{2\cdot5}=\frac{3}{2}-\frac{3}{5};\frac{9}{5\cdot8}=\frac{3}{5}-\frac{3}{8};...\)
Vậy ta có : \(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{29}-\frac{3}{32}\)
\(=\frac{3}{2}-\frac{3}{32}=\frac{3\cdot16}{32}-\frac{3}{32}=\frac{48}{32}-\frac{3}{32}=\frac{45}{32}\)
Tự so sánh , mà S đâu bé hơn 1 ???
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
b3 tính nhanh nếu có thể
a \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
b \(\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}\)
c\(1\frac{1}{24}\cdot5\frac{2}{5}\cdot2-3\frac{7}{9}\cdot2\frac{2}{17}\)
d\(2\frac{3}{13}\cdot\frac{26}{58}\cdot4\cdot2\frac{15}{24}\cdot\frac{8}{21}\)
e \(\left(1-\frac{6}{11}\right)-\frac{5}{11}\)
f\(\left(\frac{15}{7}-\frac{2}{3}\right)+\frac{2}{3}\)
g\(\left(\frac{5}{8}-\frac{1}{4}\right)+\frac{3}{8}\)
\(h\frac{3}{3\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\)
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}=\frac{31}{32}\)
b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
Tìm số nguyên âm a biết:
\(a^2-\left(\frac{3}{5}\right)^2=\frac{1}{1\cdot2}+\frac{1}{2\cdot7}+\frac{1}{7\cdot5}+\frac{1}{5\cdot13}+\frac{1}{13\cdot8}+\frac{1}{8\cdot19}+\frac{1}{19\cdot11}+\frac{1}{11\cdot25}\)