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nhung bay gio mk ban
luc nao ranh mk lam
cho nha
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\(A=\frac{8\cdot4\cdot125\cdot25+96524+2476}{10\cdot125\cdot4\cdot25\cdot8}\)
\(A=\frac{1+96524+2476}{10}\)
\(A=\frac{99001}{10}\)
\(B=\frac{5+55+555+5555}{9+99+999+9999}\)
\(B=\frac{9}{5}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)
\(=1.\frac{15}{16}=\frac{15}{16}\)
1/1x2+1/2x3+1/3x4+...+1/99x100+1
1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 +1
=1- 1/100 +1
=99/100 +1
=199/100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/999.1000 + 1
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/999 - 1/1000 + 1
= 1 - 1/1000 + 1
= 2 - 1/1000
= 2000/1000 - 1/1000
= 1999/1000
Ủng hộ mk nha ♡_♡☆_☆
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
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)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
y = \(2\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{31.31}\right)\)
3/2y =\(\frac{3}{2}.2\left(\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{31.34}\right)\)
\(\frac{3}{2}y=3\left(\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{31.34}\right)\)
\(\frac{3}{2}y=\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{31.34}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{31}-\frac{1}{34}\)]
3/2y = 1 - 1/34
3/2y = 33/34
y = 33/34 : 3/2
y =
Đúng cho mình nha
làm :
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
b, \(ab\cdot10-ab=2ab\)
\(ab\cdot10-ab\cdot1=2ab\)
\(ab\cdot\left(10-1\right)=2ab\)
\(ab\cdot9=2ab\)
\(ab\cdot9=200+ab\cdot1\)
\(ab\cdot9-ab\cdot1=200\)
\(ab\cdot\left(9-1\right)=200\)
\(ab\cdot8=200\)
\(ab=200:8\)
\(ab=25\)
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 \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)