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\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
\(\frac{1}{7}\)B=\(\frac{5}{2.7.1}+\frac{4}{1.7.11}+\frac{3}{11.2.7}+\frac{1}{2.7.15}+\frac{13}{15.4.7}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{13}{28}\)
B=\(\frac{13}{28}:\frac{1}{7}\)
B=\(\frac{13}{4}\)
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
Lương Nhất Chi
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\\ =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\\=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right) \\ =\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\)
Đừng giận nữa nha má !!!!
b) Ta có:
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}\)
| x | 7 | 9 | |||
| x2 | 49 | 81 | |||
| x2-49 | - | 0 | + | + | + |
| x2-81 | - | - | - | 0 | + |
| A | + | 0 | - | 0 | + |
dựa vào bảng ta có khi 7<x<9 thì A<0 vậy 7<x<9
b, ta có : \(\frac{2015}{1}\)+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+......+\(\frac{1}{2015}\)
=1+1+1+1......+1+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+.......+\(\frac{1}{2015}\)
(2015 số 1)
=1+(1+\(\frac{2014}{2}\))+(1+\(\frac{2013}{3}\))+........+(1+\(\frac{1}{2015}\))
=\(\frac{2016}{2016}\)+\(\frac{2016}{2}\)+\(\frac{2016}{3}\)+.........+\(\frac{2016}{2015}\)
=2016(\(\frac{1}{2016}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.........+\(\frac{1}{2015}\))
\(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
\(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{3}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Xét \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
và \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.10=\frac{10}{19}>\frac{1}{2}\)
Do đó \(B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{5}{4}>1\)
a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
Bài làm: