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A= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
B=1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110
=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11
=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1/4-1/11
=7/44
L-i-k-e nha bn hiền
A=1/1.2+1/2.3+...+1/99.100
A=1-1/2+1/2-1/3+1/3-...+1/99-1/100
A=1-1/100
A=99/100
Vậy A=99/100
Bài 1 :
Đặt A=1.2+2.3+3.4+4.5+.........+99.100
=> 3A=1.2.3+2.3.(4-1)+........+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+........+99.100.101-98.99.100
3A=99.100.101
A=33.100.101
A=333300
Bài 2 :
1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)
1)1.2+2.3+3.4+4.5+...+99.100
đặt 3D=1.2+2.3+3.4+...+99.100
=1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5
=99.100.101
=999900
D=999900:3=333300
nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2
Câu 1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{99}=\frac{49}{100}\)
cho mình nha bạn
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a,\(2\frac{2}{9}x=\frac{1}{12}+\frac{1}{20}+............+\frac{1}{72}\)
=>\(\frac{20}{9}x=\frac{1}{3.4}+\frac{1}{4.5}+.............+\frac{1}{8.9}\)
=>\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.............+\frac{1}{8}-\frac{1}{9}\)
=>\(\frac{20}{9}x=\frac{1}{3}-\frac{1}{9}\)
=>\(\frac{20}{9}x=\frac{2}{9}\)
=>x=\(\frac{1}{10}\)
b,\(\left(\frac{1}{2.3}+\frac{1}{3.4}+.............+\frac{1}{45.50}\right)x=1\)
=>\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{45}-\frac{1}{50}\right)x=1\)
=>\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
=>\(\frac{12}{25}x=1\)
=>\(x=\frac{25}{12}\)
c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)
\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)
\(\Leftrightarrow A=33\cdot100\cdot101=333300\)
b) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot25=-100\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{210}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{14.15}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{5}-\frac{1}{15}\)
Tự tính nha :)
\(B=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
Tự làm