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HT
3 tháng 7 2018
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
LT
7 tháng 7 2018
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
TV
18 tháng 9 2017
a,
x.2/7.3/4=5/21
x.3/14=5/21
x=5/21:3/14
x=10/9
b,
x.1/2=1/3
x=1/3:1/2
x=2/3
c,
x:4/5=25/8:5/4
x:4/5=5/2
x=5/2.4/5=2
KC
18 tháng 9 2017
A, x= 5/25 : 3/4 : 2/7 = 14/15
B, x=1/3 : 1/2 = 2/3
C, x=(25/8 : 5/4)x4/5 = 5/2 x 4/5 = 2
6 tháng 8 2018
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
NT
1
TH
7 tháng 8 2017
a ) \(x+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(x\times1+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(x\times1+x\times\frac{1}{3}\times\frac{9}{2}+x\times\frac{7}{2}=252\)
\(x\times1+x\times\frac{9}{6}+x\times\frac{7}{2}=252\)
\(x\times\left(1+\frac{9}{6}+\frac{7}{2}\right)=252\)
\(x\times6=252\)
\(x=252\div6\)
\(x=42\)
b ) \(\left(3\times x-0,8\right)\div x+14,5=15\)
\(\left(3\times x-0,8\right)\div x=15-14,5\)
\(\left(3\times x-0,8\right)\div x=0,5\)
\(0,8\div x=3-0,5\)
\(0,8\div x=2,5\)
\(x=0,8\div2,5\)
\(x=0,32\)
7 tháng 11 2016
a)
\(x.\frac{7}{9}=\frac{2}{3}+2\frac{1}{2}\)
\(x.\frac{7}{9}=\frac{19}{6}\)
\(x=\frac{19}{6}:\frac{7}{9}\)
\(x=\frac{57}{14}\)
b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)
\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)
\(x:\frac{9}{4}=\frac{13}{21}\)
\(x=\frac{13}{21}.\frac{9}{4}\)
\(x=\frac{39}{28}\)
12 tháng 6 2018
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
12 tháng 6 2018
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)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
TP
16 tháng 8 2017
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
TP
16 tháng 8 2017
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
DP
13 tháng 7 2017
a) \(x\cdot\frac{1}{2}+x\cdot\frac{1}{4}+x\cdot\frac{1}{8}=\frac{21}{24}\)
\(x\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=\frac{7}{8}\)
\(x\cdot\frac{7}{8}=\frac{7}{8}\)
\(\Rightarrow x=\frac{7}{8}\div\frac{7}{8}=1\)
b) \(\left(x+4\right)+\left(x+9\right)+\left(x+14\right)+.....+\left(x+44\right)+\left(x+49\right)=1430\)
\(\left(x+x+x+....+x+x\right)+\left(4+9+14+...+44+49\right)=1430\)
\(10x+265=1430\)
\(10x=1430-265\)
\(10x=1165\)
\(\Rightarrow x=\frac{1165}{10}=116,5\)
c) \(x\cdot0,25-0,5=1\)
\(x\cdot0,25=1+0,5\)
\(x\cdot0,25=1,5\)
\(\Rightarrow x=1,5\div0,25=6\)