\(\dfrac{1}{2\cdot6}\)+\(\dfrac{1}{4\cdot9}\) +\(\dfrac{1}{6\cdot12}\) +....+\(\dfrac{1}{36\cdot57}\) +\(\dfrac{1}{38\cdot60}\) dấu * là dấu nhân . mọi ng cho mình các phép tính cụ thể với ạ.
\(=\dfrac{1}{2x1x3x2}+\dfrac{1}{2x2x3x3}+\dfrac{1}{2x3x3x4}+...+\dfrac{1}{2x18x3x19}+\dfrac{1}{2x19x3x20}=\)
\(=\dfrac{1}{2x3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{18x19}+\dfrac{1}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{20-19}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}x\dfrac{19}{20}=\dfrac{19}{120}\)
Mọi người giải nhanh bài này giúp mình với, mình sắp phải nộp bài rồi😓
Thực hiện phép tính sau:
1. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
2. \(\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)
3. \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
4.\(\dfrac{x+3}{x+1}+\dfrac{2x-1}{x-1}+\dfrac{x+5}{x^2-1}\)
Câu 1: Tính giá trị biểu thức:
a.A=\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\).\(\dfrac{21}{24}\)
b.B=\(\dfrac{5}{6}\)+6\(\dfrac{5}{6}\)\(\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right)\):8\(\dfrac{1}{3}\)
c.C=1+3+6+10+15+...+1225.
Chứng tỏ:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{4}\)
(Lưu ý dấu "." trong phân số là dấu nhân)
Giúp mình với các bạn!!!
#Toán lớp 6
\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)
(Bạn nào biết thì giải giúp mình nhé, lưu ý dấu "." là dấu nhân)
Tìm x, biết:
a) -\(\dfrac{5}{6}\) - x = \(\dfrac{2}{3}\)
b) \(\dfrac{2}{3}\)x + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
c) \(\dfrac{2}{9}\) - \(\dfrac{7}{8}\) . x = \(\dfrac{1}{3}\)
d) \(\dfrac{4}{5}\) + \(\dfrac{5}{7}\) : x = \(\dfrac{1}{6}\)
e) (\(\dfrac{2}{5}\) - \(1\dfrac{2}{3}\)) : x - \(\dfrac{3}{5}\) = \(\dfrac{2}{5}\)
f) 1 - (-\(\dfrac{3}{4}\) + x) . \(2\dfrac{2}{3}\) = 0
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
\(-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}=?\)
A = -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)
= -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))
Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)
Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))
= 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))
= 2×\(\dfrac{24}{50}\)
= \(\dfrac{24}{25}\)
Thay B vào A ta có :
A = -1-\(\dfrac{24}{25}\)
=> A = \(\dfrac{-49}{25}\)
Cho mik một tick nhé thankss
B\(=-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}\)
\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)
\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)
\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)
\(\dfrac{1}{2}B=\dfrac{-49}{50}\)
\(B=\dfrac{-49}{25}\)
\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=-2*49/50
=-49/25
Chứng tỏ:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{4}\)
(Lưu ý dấu "." trong phân số là dấu nhân)
Giúp mình với các bạn!!!
Ta có:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{10.11}-\dfrac{1}{11.12}=\dfrac{1}{1.2}-\dfrac{1}{11.12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)Mà \(\dfrac{65}{132}\ne\dfrac{1}{4}\Rightarrow\) Có thể bạn ghi sai đề thì phải !
Dùng quy tắc đổi dấu rồi thực hiện các phép tính :
a) \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
b) \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
