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Trần Thùy Dung nó đã bảo \(990\ne99\cdot100\) rùi mà vẫn tách như v
=\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{4}-\frac{1}{100}\)
=\(\frac{24}{100}=\frac{6}{25}\)
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\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{24}{100}=\frac{6}{25}\)
Đặt A=1/20+1/30+1/42+1/56+...+1/930
=>A=1/4.5+1/5.6+1/6.7+...+1/30.31
=>A=1/4-1/5+1/5-1/6+...+1/30-1/31
=>A=1/4-1/31
=>A=31/124-4/124
=>A=27/124
Vây A=27/124
đề bài sai à bn phải là 1/930 chứ
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(1/2+1/4+1/8+1/16):x=1/2+1/6+1/12+.....+1/132
15/16 : x = 1/1x2+1/2x3+1/3x4+.........+1/11x12
15/16 : x = 1-1/2+1/2-1/3+1/3-1/4+........+1/11-1/12
15/16 :x = 1-1/12
15/16 : x = 11/12
x = 15/16 : 11/12
x= 45/44
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linhchi buithi, bạn ơi số 990 hình như thiếu một số 0 thì phải hay sao ý. Mình cứ thấy thiếu cái gì đó.
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{990}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}=\frac{6}{25}\)
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Ta có :
\(\frac{1}{12}=\frac{1}{12}\)
\(\frac{1}{13}< \frac{1}{12}\)
\(\frac{1}{14}< \frac{1}{12}\)
\(........\)
\(\frac{1}{17}< \frac{1}{12}\)
Cộng vế với vế ta có :
\(\frac{1}{12}+\frac{1}{13}+....+\frac{1}{17}< \frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\)(có 6 số \(\frac{1}{12}\))\(=\frac{6}{12}=\frac{1}{2}\)
Vậy \(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{17}< \frac{1}{2}\)
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1)Ta có:17=1.17=17.1=(-1).(-17)=(-17).(-1)
Do đó ta có bảng sau:
x-20 | 1 | 17 | -1 | -17 |
2y+1 | 17 | 1 | -17 | -1 |
x | 21 | 37 | 19 | 3 |
2y | 16 | 0 | -18 | -2 |
y | 8 | 0 | -9 | -1 |
Vậy các cặp (x;y) thỏa mãn là:(21;8)(37;0)(19;-9)(3;-1)
2)Ta có:7=1.7=7.1=(-1).(-7)=(-7).(-1)
Do đó ta có bảng sau:
x-10 | 1 | 7 | -1 | -7 |
y+20 | 7 | 1 | -7 | -1 |
x | 11 | 17 | 9 | 3 |
y | -13 | -19 | -27 | -21 |
Vậy các cặp (x;y) thỏa mãn là:(11;-13)(17;-19)(9;-27)(3;-21)
3)
x.y-3=12
x.y=9
Ta có:9=1.9=9.1=(-1).(-9)=(-9).(-1)=3.3=(-3).(-3)
Vậy các cặp (x;y)thỏa mãn là:(1;9)(9;1)(-1;-9)(-9;-1)(3;3)(-3;-3)
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\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)
\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{11.12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}\)
\(=\frac{1}{4}\)
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\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{50}+\dfrac{1}{990}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50}+\dfrac{1}{990}???\)
Quy luật của vế sau "..." sai, bạn xem lại đề bài!
Nếu đúng đề thì sẽ như sau:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
Đề bài đúng là như vậy.
Giải:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}\)
\(=\dfrac{25-1}{100}\)
\(=\dfrac{24}{100}\)
\(=\dfrac{6}{25}\)
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