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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 \(x=\dfrac{147}{50}\)
ND
Nguyễn Đức Trí
VIP
21 tháng 7 2023
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
NT
1
DL
22 tháng 6 2018
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}:1=5\)
\(\Rightarrow x=5-\frac{206}{100}=\frac{147}{50}\)
Vậy \(x=\frac{147}{50}.\)
TN
2
5 tháng 6 2018
Phần A sai nha phải là: 10 + 13 + ... + 79 + 82
đáp án:
SSH: A = ( 82 - 10 ) : 3 + 1 = 25
Tổng: ( 10 + 82 ) . 25 : 2 = 1150
5 tháng 6 2018
a) A = 10+13 + ...+79 + 81
A = ( 79+10) x 24 : 2 + 81
A = 89 x 24 : 2 +81
A = 1068+ 81
A= 1149
b) chỗ " 1/12.20" phải là 1/12.13 chứ !
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)
\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)
\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)
\(C=\frac{51}{91}+\frac{39}{7}\)
\(C=\frac{558}{91}\)
7 tháng 8 2016
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)
=>\(1-\frac{1}{x+1}=\frac{10}{11}\)
=>\(\frac{1}{x+1}=1-\frac{10}{11}\)
=>\(\frac{1}{x+1}=\frac{1}{11}\)
=>x+1=11
=>x=10
CM
1
14 tháng 8 2023
c: C=1*2+2*3+3*4+...+58*59+59*60
=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)
=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60
=>3*C=59*60*61
=>C=59*20*61=71980
NT
1
9 tháng 6 2016
Đặt A = 1.2 + 2.3 + 3.4 + ... + 98.99
A = 1/3 . ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3)
A = 1/3 . [ 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 98.99.(100-97)
A = 1/3 . ( 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 98.99.100 - 97.98.99)
A = 1/3 . [(1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 97.98.99)]
A = 1/3 . ( 98.99.100 - 0.1.2)
A = 1/3 .98.99.100
A = 323400
Ta có: 323400x/26950 = 12/6/7 : 3/2
12x = 14 × 2/3
12x = 28/3
x = 28/3 : 12
x = 28/3 × 1/12 = 7/9
Vậy x = 7/9
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)
\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)
\(\text{Vậy }x=64\)