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3 tháng 7 2016
b) \(x+\frac{11}{30}=\frac{19}{20}\Rightarrow x=\frac{19}{20}-\frac{11}{30}\Rightarrow x=\frac{7}{12}\)
c) \(\frac{19}{24}-x=1\Rightarrow x=\frac{19}{24}-1\Rightarrow x=-\frac{5}{24}\)
Mấy cái và và ko hỉu
t i c k cho mình 1 cái nha
BN
20 tháng 5 2018
Bài 1:
\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)
\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=\frac{9}{11}\)
Bài 2 :
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)
\(=\frac{11}{17}\)
\(=\frac{1111}{1717}=B\)
Vậy \(A=B\)
AM
13 tháng 6 2015
1)\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)
13 tháng 6 2015
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
=> \(A<\frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}=\frac{11}{17}=\frac{1111}{1717}=B\)
Vậy A < B
XO
16 tháng 8 2019
\(\left(\frac{2}{31\times33}+\frac{2}{33\times35}+...+\frac{2}{39\times41}\right)\times2542-2,04:x=19\)
\(\Rightarrow\left(\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}+...+\frac{1}{39}-\frac{1}{41}\right)\times2542-2,04:x=19\)
\(\Rightarrow\left(\frac{1}{31}-\frac{1}{41}\right)\times2542-2,04:x=19\)
\(\Rightarrow\frac{10}{1271}\times2542-2,04:x=19\)
\(\Rightarrow20-2,04:x=19\)
\(\Rightarrow2,04:x=20-19\)
\(\Rightarrow2,04:x=1\)
\(\Rightarrow x=2,04:1\)
\(\Rightarrow x=2,04\)
Vậy x = 2,04
S
16 tháng 8 2019
\(\left(\frac{2}{31\text{ x }33}+\frac{2}{33\text{ x }35}+...+\frac{2}{39\text{ x }41}\right)\text{ x }2542-2,04\text{ : }x=19\)
\(\left(\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}+...+\frac{1}{39}-\frac{1}{41}\right)\text{ x }2542-2,04\text{ : }x=1\)
\(\left(\frac{1}{31}-\frac{1}{41}\right)\text{ x }2542-2,04\text{ : }x=1\)
\(\frac{10}{1271}\text{ x }2542-2,04\text{ : }x=1\)
\(20-2,04\text{ : }x=1\)
\(2,04\text{ : }x=20-1\)
\(2,04\text{ : }x=19\)
\(x=2,04\text{ : }19\)
\(x=\frac{51}{475}\)
9 tháng 8 2015
a) = 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35
b) = 1/3 x 4/5 + 1/3 x6/5 + 1/3 x 2 = 1/3(4/5 + 6/5 + 2) = 1/3 x 4 = = 4/3
c) 4/7 x 2/9 + 4/7 x 7/9 + 2/3 = 4/7 x (2/9 + 7/9) + 2/3 = 4/7 x 1 + 2/3 = 26/21
4 tháng 3 2022
A) 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35
B) 1/3 x 4/5 + 1/3 x 6/5 + 1/3 x 2 = 1/3 x(4/5 + 6/5 x 2 ) = 1/3 x 4 = 4/3
c) TƯƠNG TỰ CÂU A VÀ B
* HOKTOT*
NHA
LP
2
12 tháng 10 2016
a) \(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)
\(x:\frac{1}{2}+\frac{3}{4}=\frac{39}{20}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)
\(x:\frac{1}{2}=\frac{24}{20}\)
\(x=\frac{24}{20}.\frac{1}{2}\)
\(x=\frac{3}{5}\)
AY
12 tháng 10 2016
\(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)
\(x:\frac{1}{2}=1\frac{19}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)
\(x:\frac{1}{2}=\frac{24}{20}\)
\(x=\frac{24}{20}\times\frac{1}{2}\)
\(x=\frac{24}{40}\)
\(x=\frac{3}{5}\)
1 tháng 9 2020
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
NS
0