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\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
<=> \(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
<=> \(94< x< 92\)vô lí
Vậy không tìm đc x thỏa mãn
\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(=\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow94< x< 92\)
\(\Rightarrow\)ĐỀ SAI
có thể đây là bài lớp 4 nhưng mình nghĩ là các bạn lớp 5 cũng sẽ khó khăn đó
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{50.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2550}\right)\)
\(=\frac{1}{2}.\left(\frac{1275}{2550}-\frac{1}{2550}\right)\)
\(=\frac{1}{2}.\frac{1274}{2550}\)
\(=\frac{637}{2550}\)
Lưu ý : Dấu \("."\)là dấu \("\)x \("\)
( dấu nhân )
Chúc bạn học giỏi !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
VD ( dễ hiểu )
Ta có: \(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{6}-\frac{1}{11}=\frac{5}{66}\)
\(\Rightarrow M=\frac{5}{66}:\frac{1}{2}=\frac{5}{33}.\)
\(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(M=\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}\)
\(M=\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}+\frac{2}{10\cdot11}\)
\(M=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)
\(M=2\left(\frac{1}{6}-\frac{1}{11}\right)\)
\(M=2\cdot\frac{5}{66}\)
\(M=\frac{5}{33}\)
\(8\frac{1}{6}:2\frac{1}{2}=\frac{49}{6}:\frac{5}{2}=\frac{98}{30}=\frac{49}{15}\)
\(1\frac{3}{10}:5\frac{7}{8}=\frac{13}{10}:\frac{47}{8}=\frac{104}{470}=\frac{52}{235}\)
Bn như thần ý nhỉ :)
trả lời:
8x1/6:2x1/2
=4/3:1
=4/3
1x3/10:5x7/8
=3/10:35/8
=12/175
chúc học tốt
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)
Do not ask why hay quá!
Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)
Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)
\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)
Vậy .. . .
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
Sửa đề chút:
\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)
\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{4}-\frac{1}{99.200}< 1\)
đpcm
thank you nha!!!