tính
a, \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+......+\dfrac{1}{9500}\)
b, \(\dfrac{3}{2}-\dfrac{5}{6}-\dfrac{7}{12}-\dfrac{9}{20}-.....-\dfrac{19}{90}\)
Chứng minh rằng: \(\dfrac{3}{2^2}+\dfrac{5}{6^2}+\dfrac{7}{12^2}+\dfrac{9}{20^2}+...+\dfrac{19}{90^2}< 1\)
=> 1 - 1 /2^2 + 1 /2^2 -1 /3^2 + 1/3^2 - 1/4^2 + .... + 1/9^2 - 1/10^2 <1 => 1 - 1/10^2 <1 ( luôn đúng )
Tính nhanh:
a, \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
b, \(\left(-\dfrac{1}{4}+\dfrac{7}{35}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
Tính (theo mẫu).
Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
I = \(\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}\)
J = \(\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}\)
K = \(\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}\)
L = \(\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}\)
\(I=\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}=\dfrac{9}{8}\)
\(J=\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}=\dfrac{239}{99}\)
\(K=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}=-\dfrac{13}{6}\)
\(L=\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}=\dfrac{103}{60}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
Bài1. (4điểm) Thực hiện phép tính:
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(=\dfrac{3}{5}+\dfrac{41}{25}\)
\(=\dfrac{15}{25}+\dfrac{41}{25}\)
\(=\dfrac{56}{25}\)
b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\) \(1-\dfrac{1}{10}\)
\(=\dfrac{-9}{10}\)
Tính hợp lý nếu có thể:
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
c) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}\)
d) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}\)
a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
a,\(=\dfrac{2}{9}+\left(\dfrac{-3}{10}+\dfrac{-7}{10}\right)=\dfrac{2}{9}+\left(-1\right)=\dfrac{-7}{9}\)
b,\(=\left(\dfrac{-11}{6}+\dfrac{-1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-8}{5}\)
c,\(=\dfrac{27}{13}-\left(\dfrac{106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
VÒNG 2
Bài 1: Mèo con nhanh nhẹn
\(\dfrac{1}{2}\) + \(\dfrac{1}{12}\) | 2 + \(\dfrac{1}{6}\) | \(\dfrac{1}{20}\) | 1 - \(\dfrac{1}{9}\) | |
\(\dfrac{1}{15}\) + \(\dfrac{2}{15}\) | \(\dfrac{1}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{7}{12}\) | \(\dfrac{4}{12}\) | |
\(\dfrac{9}{14}\)+ \(\dfrac{1}{14}\) | 1 + \(\dfrac{1}{6}\) | \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) | \(\dfrac{1}{3}\) - \(\dfrac{2}{9}\) | |
\(\dfrac{3}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{1}{5}\) | 1 - \(\dfrac{8}{9}\) | ||
\(\dfrac{5}{7}\) | 1 - \(\dfrac{2}{3}\) | \(\dfrac{1}{3}\) + \(\dfrac{5}{9}\) |
Chứng tỏ rằng: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)