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DM
16 tháng 3 2017
Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)
Quy đồng mẫu số :
\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)
\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)
Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)
HN
1
25 tháng 6 2022
=>360+57<10x<58x4+59x5
=>417<10x<527
\(\Leftrightarrow10x\in\left\{420;430;440;...;510;520\right\}\)
hay \(x\in\left\{42;43;44;...;51;52\right\}\)
Q
10 tháng 4 2017
\(Q=\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)
\(Q=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)+\left(1-\dfrac{1}{72}\right)+\left(1-\dfrac{1}{90}\right)\)
\(Q=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(Q=9-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(Q=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(Q=9-\left(1-\dfrac{1}{10}\right)\)
\(Q=9-\dfrac{9}{10}\)
\(Q=\dfrac{81}{10}\)
Chúc bạn học tốt :))
CL
29 tháng 8 2017
\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+1-\dfrac{1}{30}+1-\dfrac{1}{42}+1-\dfrac{1}{56}+1-\dfrac{1}{72}+1-\dfrac{1}{90}\)
\(=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=9-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=9-\left(1-\dfrac{1}{10}\right)\)
\(=9-\dfrac{9}{10}\)
\(=\dfrac{81}{10}\)
EJ
14 tháng 8 2017
a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
JP
14 tháng 8 2017
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)
KD
1
TM
28 tháng 3 2017
Ta có:
A = \(\dfrac{100^{100}+1}{100^{90}+1}\)> 1 \(\Rightarrow\) A > \(\dfrac{100^{100}+1+99}{100^{90}+1+99}\) = \(\dfrac{100^{100}+100}{100^{90}+100}\)
\(\Rightarrow\) A > \(\dfrac{100\left(100^{99}+1\right)}{100\left(100^{89}+1\right)}\) = B
Vậy A > B
17 tháng 4 2017
a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
C
17 tháng 4 2017
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)