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\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+........+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(=2\cdot\frac{9}{10}=\frac{9}{5}\)
đúng nha !
Ta có:
2/1.2+2/2.3+2/3.4+2/4.5+2/5.6+2/6.7+2/7.8+2/8.9+2/9.10
=2.(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)
=2.(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=2.(1-1/10)
=2.9/10
=9/5
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}\right)\cdot x=\dfrac{23}{45}\)
\(\Leftrightarrow x\left(1-\dfrac{1}{9}\right)=\dfrac{23}{45}\)
\(\Leftrightarrow x=\dfrac{23}{45}\cdot\dfrac{9}{8}=\dfrac{23}{40}\)
Đặt A bằng biểu thức trong ngoặc
\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{10-8}{8.9.10}\)
\(2A=\dfrac{1}{2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{9.10}\)
\(2A=\dfrac{44}{90}\)
\(A=\dfrac{22}{90}\)
Lời giải:
$A=1.2+2.3+3.4+...+8.9+9.10$
$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+8.9(10-7)+9.10(11-8)$
$=(1.2.3+2.3.4+3.4.5+...+8.9.10+9.10.11)-(0.1.2+1.2.3+2.3.4+...+8.9.10)$
$=9.10.11$
$\Rightarrow A=\frac{9.10.11}{3}=330$
E= 1.2 + 2.3 + 3.4 +...+ 8.9 + 9.10=2+6+12+20+30+42+56+72+90=330 lộn
\(=1-2\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=1-2\cdot\dfrac{7}{18}=\dfrac{18}{18}-\dfrac{14}{18}=\dfrac{4}{18}=\dfrac{2}{9}\)