Bài 3 tính nhanh a,5/6-1/2-1/4-1/8-1/16 b,1/3+1/6+1/12+1/24+1/48+1/96 c,1/1x2+1/2x3+1/3x4+1/4x5+1/5x6 d,1/2+1/6+1/121/20+1/30+1/42 khẩn cấp aaaaa..... huhuhu.....
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1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)
\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)
\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)
1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)
\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)
\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)
\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)
2) a) \(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}\)
\(x=\frac{35}{9}x\frac{6}{35}\)
\(x=\frac{2}{3}\)
b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)
\(\frac{5}{6}x10-X=0\)
\(X=\frac{5}{6}x10=\frac{25}{3}\)
Đúng nha !!!!
1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)
b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)
c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)
2/a/\(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)
\(x=\frac{2}{3}\)
b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)
\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)
\(\frac{47}{60}\cdot10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
1.
a.
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
b.
Tích có 100 thừa số
=> n = 100
\(\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times\left(100-100\right)\)
\(=\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times0\)
\(=0\)
2.
a.
\(135\times789789-789\times135135=1001\times\left(135\times789-789\times135\right)=1001\times0=0\)
b.
\(\left(28\times9696-96\times2828\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left[101\times\left(28\times96-96\times28\right)\right]\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left(101\times0\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\)
3.
a.
\(\left[\left(x+32\right)-17\right]\times2=42\)
\(\left(x+32\right)-17=\frac{42}{2}\)
\(\left(x+32\right)-17=21\)
\(x+32=21+17\)
\(x+32=38\)
\(x=38-32\)
\(x=6\)
b.
\(125+\left(145-x\right)=175\)
\(145-x=175-125\)
\(145-x=50\)
\(x=145-50\)
\(x=95\)
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)