15/4+15/28+15/70+..+15/x+(x+3)=99/20
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a) Ta có B = \(\left(\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+\frac{6}{391}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\frac{20}{69}.x.\left(x-1\right)=\frac{20}{69}\)
=> \(x.\left(x-1\right)=\frac{20}{69}:\frac{20}{69}\)
=> \(x.\left(x-1\right)=1\)
=> \(x\in\varnothing\)
a) \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)
=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)
=> \(\frac{31}{94}.x=\frac{31}{94}\)
=> \(x=\frac{31}{94}:\frac{31}{94}\)
=> \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Ta có : ( 1 + 2 + 3 + ... + 99)
Số số hạng là: ( 99 - 1 ) : 1 + 1 = 100
Tổng là: ( 99 + 1 ) x 100 : 2 = 5000
=> 5000 x ( 13 - 12 - 1 ) x 15
=> 5000 x 10 x 15
=> 50000 x 15
=> 750000
Ko muốn vt nx :))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4/15 : 4/7 < x < 2/5 x 10/3
7/15 < x < 4/3
Vậy số tự nhiên x chỉ có thể là 1 nha .
\(\frac{15}{4}+\frac{15}{28}+\frac{15}{70}+...+\frac{15}{x.\left(x+3\right)}=\frac{99}{20}\)
=> \(5.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{99}{20}\)
=> \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{99}{20}:5\)
=> \(1-\frac{1}{x+3}=\frac{99}{20}.\frac{1}{5}=\frac{99}{100}\)
=> \(\frac{1}{x+3}=1-\frac{99}{100}\)
=> \(\frac{1}{x+3}=\frac{1}{100}\)
=> x + 3 = 100
=> x = 100 - 3
=> x = 97