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\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)
\(1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times1\frac{1}{6}\times1\frac{1}{7}\times1\frac{1}{8}\)
\(=\)\(\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times\frac{9}{8}\)
\(=\)\(\frac{9}{3}\)
\(=\)\(3\)
\(1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times1\frac{1}{6}\times1\frac{1}{7}\times1\frac{1}{8}\)
\(=\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times\frac{9}{8}\)
Rút gọn phép tính trên ta được :
\(=\frac{9}{3}=3\)
\(3\frac{3}{4}\times4=\frac{3\times4+3}{4}\times4=\frac{15}{4}\times4=\frac{15\times4}{4}=15\)
\(4\frac{3}{2}\div1\frac{1}{2}=\frac{4\times2+3}{2}\div\frac{1\times2+1}{2}=\frac{11}{2}\div\frac{3}{2}=\frac{11}{2}\times\frac{2}{3}=\frac{11}{3}\)
\(4\frac{3}{4}+2\frac{3}{2}=\frac{4\times4+3}{4}+\frac{2\times2+3}{2}=\frac{19}{4}+\frac{7}{2}=\frac{19}{4}+\frac{14}{4}=\frac{33}{4}\)
\(3\frac{8}{6}-4\frac{1}{3}=\frac{3\times8+6}{6}-\frac{4\times3+1}{3}=\frac{30}{6}-\frac{13}{3}=\frac{30}{6}-\frac{26}{6}=\frac{4}{6}=\frac{2}{3}\)
\(3\frac{2}{4}\)x 4 = \(\frac{14}{4}\)x 4 = \(\frac{56}{4}\)
\(4\frac{3}{2}\): \(1\frac{1}{2}\)= \(\frac{11}{2}\): \(\frac{3}{2}\)= \(\frac{11}{2}\)x \(\frac{2}{3}\)= \(\frac{22}{6}\)= \(\frac{11}{3}\)
\(4\frac{3}{4}\)+ \(2\frac{3}{2}\)= \(\frac{11}{4}\)+ \(\frac{7}{2}\)= \(\frac{25}{4}\)= 6,25
\(3\frac{8}{6}\)- \(4\frac{1}{3}\)= \(\frac{13}{3}\)- \(\frac{13}{3}\)= 0
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
\(=\frac{6}{5}\times\frac{7}{6}\times...\times\frac{11}{10}\)(lại lỗi đề)
\(=\frac{6×7×...×11}{5×6×...×10}\)
\(=\frac{11}{5}\)
\(1\frac{1}{5}\cdot1\frac{1}{6}\cdot1\frac{1}{7}\cdot1\frac{1}{8}\cdot1\frac{1}{9}\cdot1\frac{1}{10}\)
\(=\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}\cdot\frac{9}{8}\cdot\frac{10}{9}\cdot\frac{11}{10}\)
\(=\frac{6\cdot7\cdot8\cdot9\cdot10\cdot11}{5\cdot6\cdot7\cdot8\cdot9\cdot10}\)
\(=\frac{11}{5}\)
Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))
2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)
A = \(\frac{728}{729}:2=\frac{364}{729}\)
a.5/6 - 26/5 X 1/13 = 13/30
b.( 19/23 - 22/46 ) X 23/46 = 3/23
c.25/8 x 14/30 = 35/24
d.( 3/4 x 5/7 ) x ( 20/9 x 14/15 ) = 10/9
e.4/35 x 25/32 x 38/24 = 95/672
g. 1/2 x 3/4 x 2/3 x 4/5 = 1/5
h.5/6 x 11/4 - 5/4 x 23/6 = -5/2
i.9/16 x 13/4 - 9/4 x 5/16 + 9/16 x 17/4 = 225/64
k.( 7 x 1/3 ) x ( 1/7 x 6 ) = 2
m.2/3 x ( 3/5 + 3/7 ) = 34/35
n.4/5 x ( 5/8 + 7/4 ) = 19/10
p.( 1/33 + 31/333 - 341/3333 ) x ( 1/2 - 1/3 - 1/6 ) = 0
mih giành cả nửa tiếng để giải đó , k nha
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
ko ghi lại đề
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)
\(=1-\frac{1}{212}\)
\(=\frac{211}{212}\)
\(\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{211.212}\)
= \(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{211}-\frac{1}{212}\)
= \(\frac{1}{8}-\frac{1}{212}\)
= \(\frac{51}{424}\)