\(\frac{6}{2x5}+\frac{6}{5x8}+\frac{6}{8x11}+.......+\frac{6}{98x101}\)
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Giải:
Ta có:
\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)
\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)
\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
_Chúc bạn học tốt_
Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)
\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)
\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)
\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)
\(3xA=\frac{1}{2}-\frac{1}{20}\)
\(3xA=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}\)
\(=\frac{3}{7}\)
3/2x5 + 3/5x8 + 3/8x11 + 3/11x14
= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14
= 3/2 - 3/14
= 21/14 - 3/14
= 18/14
= 9/5
\(\frac{9}{2.5}+\frac{39}{5.8}+\frac{87}{8.11}+...+\frac{9897}{98.101}\)
\(=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{98.101}\)
\(=1+1+...+1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\right)\) \(\left(\text{33 chữ số 1}\right)\)
\(=33-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\right)\)
\(=33-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=3-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=3-\frac{1}{3}-\frac{99}{202}\)
\(=\frac{1319}{606}\)
\(\frac{2}{2x5}+\frac{2}{5x8}+\frac{2}{8x11}+...+\frac{2}{96x98}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{98}\right)\)
\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{2}{3}x\frac{24}{49}=\frac{16}{49}\)
= 2 x ( 1/2 x 5 + 1/ 5 x 8 + 1/ 8 x 11 + 1/ 11 x 14 + 1/ 14 x 17 )
= 2 x ( 1/2 - 1/5 + 1/5 - 1/8 + ....+1/14 - 1/17)
= 2 x (1/2 - 1/17)
= 2 x 15/34
= 15/17
ĐÚNG THÌ TÍCH CHO MÌNH NHA
CHÚC BẠN HỌC GIỎI
Đặt \(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\right)\)
\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{17}\right)\)
\(A=\frac{2}{3}\cdot\frac{15}{34}=\frac{5}{17}\Rightarrow A< 1\)
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
Đặt \(Shin=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)
\(\Rightarrow3Shin=\frac{6}{2}-\frac{6}{5}+\frac{6}{5}-\frac{6}{8}+\frac{6}{8}-\frac{6}{11}+...+\frac{6}{98}-\frac{6}{101}\)
\(\Leftrightarrow3Shin=\frac{6}{2}-\frac{6}{101}=\frac{297}{101}\)
N/t: . là dấu nhân nha! Cái đó lớp 5 chưa biết đâu! Lên cấp 2 mới học. Trong bài làm bạn cứ ghi là dấu " x" thay cho dấu "." của mình nha!
Đặt A = 6/2.5 + 6/5.8 + ... + 6/98.101
=> A = 6.(1/2.5 + 1/5.8 + ... + 1/98.101)
=> 3A = 6.(3/2.5 + 3/5.8 + ... + 3/98.101)
=> 3A = 6.(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/98 - 1/101)
=> 3A = 6.99/202
=> 3A = 297/101
=> A = 99/101