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\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
b, \(ab\cdot10-ab=2ab\)
\(ab\cdot10-ab\cdot1=2ab\)
\(ab\cdot\left(10-1\right)=2ab\)
\(ab\cdot9=2ab\)
\(ab\cdot9=200+ab\cdot1\)
\(ab\cdot9-ab\cdot1=200\)
\(ab\cdot\left(9-1\right)=200\)
\(ab\cdot8=200\)
\(ab=200:8\)
\(ab=25\)
\(A=\frac{8\cdot4\cdot125\cdot25+96524+2476}{10\cdot125\cdot4\cdot25\cdot8}\)
\(A=\frac{1+96524+2476}{10}\)
\(A=\frac{99001}{10}\)
\(B=\frac{5+55+555+5555}{9+99+999+9999}\)
\(B=\frac{9}{5}\)
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
=\(\frac{6\left(1+8+27+64\right)}{12\left(1+16+54+128\right)}\)
=\(\frac{6.100}{12.199}\)
=\(\frac{50}{199}\)
Tk mình với nha mọi người!!!!!
\(\frac{1x2x3+2x4x6+3x6x9+4x8x12}{1x3x4+4x6x8+6x9x12+8x12x16}\)
\(\frac{6x\left(1+8+27+64\right)}{12x\left(1+16+54+128\right)}=\frac{6x100}{12x199}=\frac{50}{199}\)
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
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 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
Gọi \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=1-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=\frac{24}{25}\)
\(\Leftrightarrow\)\(A=\frac{24}{25}:3\)
\(\Leftrightarrow\)\(A=\frac{24}{25}.\frac{1}{3}\)
\(\Leftrightarrow\)\(A=\frac{8}{25}\)
Vậy \(A=\frac{8}{25}\)
Đặt \(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)
\(\Rightarrow3C=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{22.25}\)
\(\Rightarrow3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Rightarrow3C=1-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow C=\frac{24}{25}:3=\frac{8}{25}\)
Vậy \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)