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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
A ) 2/17 + 30/46 + 15/17 + 4/49 + 8/23=102/49
B ) 5/8 x ( 19/31 x 8/5 )=19/31
C ) 7/9 x 8/15 x 9/7 x 15/4=2
D ) 11/12 x 24/7 x 21/22=3
nhìu quá nên mk chỉ cho kết quả thôi nha ^-^!!!
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
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
a)\(\frac{27}{5}\text{x}\frac{91}{12}\text{x}\frac{125}{9}\text{x}\frac{96}{13}=\frac{27}{9}\text{x}\frac{91}{13}\text{x}\frac{125}{5}\text{x}\frac{96}{12}=3\text{x}7\text{x}25\text{x}8\)
Đến đây tự tính nha !
b)\(\frac{2539}{35}\text{x}\frac{7}{90}+\frac{561}{35}\text{x}\frac{7}{90}=\frac{7}{90}\text{x}\left(\frac{2539}{35}+\frac{561}{35}\right)=\frac{7}{90}\text{x}\frac{3100}{35}=\frac{3100}{90}\text{x}\frac{7}{35}=\frac{310}{9}\text{x}\frac{1}{5}=\frac{62}{9}\)
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
Tính :
a) 13/15 + 4/7 - 101/105 = 91/105 + 60/105 - 101/105
= 151/105 - 101/105
= 50/105
= 10/21
b) 2/5 * 3/5 * 4/9
= 15/25 * 4/9
= 60/225
Tìm x :
a) x + 11/4 = 17/3 b) x = 9/5 - 23/7
x = 17/3 - 11/4 x = 23/7 - 9/5
x = 35/12 x = 52/35
c) x * 7/2 = 19/4 d) x : 8/3 = 13/3
x = 19/4 : 7/2 x = 13/3 * 8/3
x = 38/28 x = 104/9
x = 19/14
Ủng hộ nha !!!
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\)
\(=\dfrac{7-3}{3\times7}+\dfrac{11-7}{7\times11}+\dfrac{15-11}{11\times15}+\dfrac{19-15}{15\times19}+\dfrac{23-19}{19\times23}+\dfrac{27-23}{23\times27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}=\dfrac{8}{27}\)