(1/1*51+1/2*52+1/3*53+...+1/10*60)x=1/1*11+1/2*12+1/3*13+...+1/50*60
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HB
0
PJ
7 tháng 7 2019
a) 50 + 48 + 46 + ... + 4 - 47 - 45 - 43 - ... - 1
= (50 - 45) + (48 - 43) + (46 - 41) + ... + (6 - 1) + (4 - 47)
=72
Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 5
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + ... + 50 - 51 - 52 + 53 + 54
= (1 + 54) + (2 + 53) - (3 + 52) - (4 + 51) + ... + (25 + 30) + (26 + 29) - (27 + 28)
=55
Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 55. Còn dấu đằng trước nhóm thì theo dấu đề bài cho
~ Học tốt ~
PT
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 5 2023
(1 + \(\dfrac{1}{49}\))\(\times\)(1 + \(\dfrac{1}{50}\))\(\times\)(1 + \(\dfrac{1}{51}\))\(\times\)(1 + \(\dfrac{1}{52}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{60}\))
= \(\dfrac{49+1}{49}\) \(\times\) \(\dfrac{50+1}{50}\)\(\times\) \(\dfrac{51+1}{51}\)\(\times\)\(\dfrac{52+1}{52}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)
= \(\dfrac{50}{49}\)\(\times\)\(\dfrac{51}{50}\)\(\times\)\(\dfrac{52}{51}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)
= \(\dfrac{50\times51\times52\times53\times...\times60}{50\times51\times52\times53\times...\times60}\)\(\times\)\(\dfrac{61}{49}\)
= \(\dfrac{61}{49}\)
BT
2 tháng 6 2017
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
S
29 tháng 4 2017
a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
VT
1
T
0