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`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)
`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)
`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)
`= -21/24 = -7/8`
`b)`
\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)
`c)`
\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)
`d) `
\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)
`e)`
\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)
`= 17*(23-53+14-24)`
`= 17*(-40)`
`= -680`
`f)`
\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)
`= 19*(-218-82-533-167)`
`= 19*(-1000)`
`= -19000`
`g)`
\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)
`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)
`h)`
\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)
`i )`
\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)
`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)
`k)`
\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)
`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)
`l )`
\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)
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`# \text {KaizulvG}`
a) 1/2 + 3/4 - (3/4 - 4 - 5)
= 1/2 + 3/4 - 3/4 + 4 + 5
= (3/4 - 3/4) + (4 + 5) + 1/2
= 0 + 9 + 1/2
= 19/2
b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)
= 9/16 - 8/27 - 19/27 + 7/16 + 2
= (9/16 + 7/16) + (-8/27 - 19/27) + 2
= 1 - 1 + 2
= 2
c) -5/8 . [4/9 + 7/(-12)]
= -5/8 . (-5/36)
= 25/288
d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)
= 7/10 . (-3/5 - 2/5) + 3/10
= 7/10 . (-1) + 3/10
= -2/5
e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7
= -3/7 . (5/9 + 4/9) + 17/7
= -3/7 . 1 + 17/7
= 2
f) 8 2/7 - (3 4/9 + 4 2/7)
= 8 + 2/7 - 3 - 4/9 - 4 - 2/7
= (8 - 3 - 4) + (2/7 - 2/7) - 4/9
= 1 - 4/9
= 5/9
h) 3.(-1/2)² - (4/5 + 8/15) : 5/6
= 3.1/4 - 4/3 : 5/6
= 3/4 - 8/5
= -17/20
A=1+2+3+4+5+...+50
A=(50+1)+(49+2)+(48+3)+...
A=(50+1)*[(50-1):1+1]:2
A=51*25=1275
B=2+4+6+8+10+...+100
B=(100+2)+(98+4)+(96+6)+...
B=(100+2)*[(100-2):2+1]:2
B=102*25=2550
C=1+4+7+10+13+...+99
C=(99+1)+(96+4)+(93+7)+...
C=(99+1)*[(99-1):3+1]:2
C=100*16.8333=1683.33
D=2+5+8+11+14+...+98
D=(98+2)+(95+5)+(92+8)+...
D=(98+2)*[(98-2):3+1]:2
D=100*16.5=1650
E=1+2+3+4+5+...+25
E=(25+1)+(24+2)+(23+3)+...
E=(25+1)*[(25-1):1+1]:2
E=26*12.5=325
F=2+4+6+8+10+...+50
F=(50+2)+(48+4)+(46+6)+...
F=(50+2)*[(50-2):2+1]:2
F=52*12.5=650
G=3+5+7+9+11+...+51
G=(51+3)+(49+5)+(47+7)+...
G=(51+3)*[(51-3):2+1]:2
G=54*12.5=675
H=1+5+9+13+17+...+81
H=(81+1)+(77+5)+(73+9)+...
H=(81+1)*[(81-1):4+1]:2
H=82*10.5=861
a) A =1 + 2 + 3 + 4 + … + 50
Số số hạng của dãy số trên là:
(50 - 1) : 1 + 1 = 50 (số số hạng)
A =(1+ 50) . 50 : 2
= 51 . 50 : 2
= 2550 : 2
= 1275
b) B = 2 + 4 + 6 + 8 + ... + 100
Số số hạng của dãy số trên là:
(100 - 2) : 2 + 1 = 50 (số hạng)
Có số cặp là:
50 : 2 = 25 (cặp)
Tổng của 1 cặp là:
100 + 2 = 102
Tổng của dãy số là:
25 .102 = 2550
c) C = 1 + 3 + 5 + 7 + … + 99
Số số hạng của dãy trên là:
(99 - 1) : 2 + 1 = 50 (số số hạng)
C = (1 + 99) . 50 : 2
= 100 . 50 : 2
= 5000 : 2
= 2500
d) D = 2 + 5 + 8 + 11 + … + 98
Số số hạng của dãy trên là:
(98 - 2) : 3 + 1 = 33 (số số hạng)
=> Dãy trên có 16 cặp
D = (95 + 2) .16 + 98
= 97 . 16 + 98
= 1552 +98
= 1650
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
B=4*13/9*3-4/3*40/9
B=4/3*13/9-4/3*40/9
B=4/3*(13/9-40/9)
B=4/3*(-27)/9
B=4*(-3)/9
B=-4
A=6/7 + 1/7.(2/7+5/7)
A=6/7 + 1/7.7/7=6/7+1/7.1
A=6/7+1/7=7/7=1
d)\(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) 11/4x(-0,4)x-1,6x11/4
= -1,1 x-1,6x11/4
= 1,76x11/4
= 4,48
b) 6/19x-7/11+6/19x-4/19x-4/11+13/19
= -42/209+96/3971+13/19
= -702/3971+13/19
= 2015/3971
c) 2/5x3/7+-3/7x3/5+3/7
= 3/7x(2/5+3/5+1)
= 3/7x(1+1)
= 3/7x2
= 6/7
d) 2/3x5+2x5/7+2/7x9+....+2/97x99
= 1/3-1/5+1/5-1/7+1/7-1/9+......+1/97-1/99
= 1/3-1/99
= 32/99
b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)
c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)
d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)
e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)