tim x biet
5/1.2 + 5/2.3 + 5/3.4 + …. 5/x(x+1) = 9998/9999
K
Khách
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10 tháng 9 2017
1) (x-3)(x-5) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
(x+7).35 = 2.35
\(\Rightarrow x+7=2\)
\(\Rightarrow x=2-7=-5\)
Vậy x = -5
2) 1.2 + 2.3 + 3.4 + .... + 99.100
Đặt A = 1.2 + 2.3 + .... + 99.100
3A = 1.2.3 + 2.3.4 + 3.4.3 + .... + 99.100.3
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)+ .... + 99.100.(101-98)
3A = ( 1.2.3 + 2.3.4 + 3.4.5 +.... + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 98.99.100 )
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A = 999900
A = 999900 : 3
A = 333300
Vậy A = 333300
DP
31 tháng 7 2017
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)
\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)
\(\text{Vậy }x=64\)
TN
0
1 tháng 8 2017
Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)
\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x + 1 = 100
=> x = 99
LP
1
20 tháng 8 2021
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{103}{50}=5\)
hay \(x=\dfrac{147}{50}\)
NT
1
HN
15 tháng 7 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
DT
2
24 tháng 6 2015
Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE
K
24 tháng 6 2015
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)
\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)
\(3.\frac{99}{100}+4x=2500\)
\(\frac{297}{100}+4x=2500\)
\(4x=2500-\frac{297}{100}\)
\(4x=2500-2,97\)
\(4x=2497,03\)
\(x=624,2575\)
\(x=2497,03:4\)
NN
0
KW
1
KN
7 tháng 3 2016
HD;Nhìn chung thấy có (x.99.100) cuối cùng => có 99 x Và tách ra ta đc 1.2.3......99.100+99x nha! 2
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{9998}{9999}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{9998}{9999}\)
\(\Leftrightarrow5\left(1-\frac{1}{x+1}\right)=\frac{9998}{9999}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{9999}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{49995}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9998}{49995}=\frac{39997}{49995}\)
\(\Leftrightarrow x=\frac{9998}{39997}\)