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\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
a: \(\left(-17\right)+5+\left(-8\right)+17\)
\(=\left(-17+17\right)+\left(5-8\right)\)
=0-3
=-3
b: \(\left(-9\right)+\left(-11\right)+21+\left(-1\right)\)
\(=\left(-9-1\right)+\left(21-11\right)\)
=10-10
=0
c: \(15+23+\left(-25\right)+\left(-23\right)\)
\(=\left(15-25\right)+\left(23-23\right)\)
=-10+0
=-10
d: \(\left(-3\right)+\left(-350\right)+\left(-7\right)+350\)
\(=\left(-3-7\right)+\left(-350+350\right)\)
=-10+0
=-10
e: \(\left(-3\right)+\left(-250\right)+\left(-7\right)+250\)
\(=\left(-3-7\right)+\left(-250+250\right)\)
=-10+0
=-10
f: \(\left(-4\right)+\left(-440\right)+\left(-6\right)+440\)
\(=\left(-4-6\right)+\left(-440+440\right)\)
=-10+0
=-10
g: \(483+\left(-56\right)+263+\left(-64\right)\)
\(=\left(483+263\right)+\left(-56-64\right)\)
\(=746-110=636\)
h: \(215+43+\left(-215\right)+\left(-25\right)\)
\(=\left(215-215\right)+\left(43-25\right)\)
=0+18
=18
a. (-17) + 5 + (-8) + 17
= [5 + (-8)] + [(-17) + 17]
= -3
b. (-9) + (-11) + 21 + (-1)
= [(-9) + (-1)] + [21 + (-11)]
= -10 + 10
= 0
c. 15 + 23 + (-25) + (-23)
= [15 + (-25)] + (23 - 23)
= -10
d. (-3) + (-350) + (-7) + 350
= [(-3) + (-7)] + [(-350) + 350]
= -10
e. (-3) + (-250) + (-7) + 250
= [(-3) + (-7)] + [(-250) + 250]
= -10
f. (-4) + (-440) + (-6) + 440
= [(-4) + (-6)] + [(-440) + 440]
= -10
g. 483 + (-56) + 263 + (-64)
= (483 + 263) + [(-56) + (-64)]
= 746 + (-120)
= 626
h. 215 + 43 + (-215) + (-25)
= [215 + (-215)] + [43 + (-25)]
= 18
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
a) \(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{97}{56}\)
b) \(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)
c) \(\dfrac{2}{3}.\dfrac{4}{9}=\dfrac{8}{27}\)
d) \(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{7}{10}\)
a: =48/56+49/56
=97/56
b: =12/15-10/15
=2/15
c: =(2*4)/(3*9)=8/27
d: =1/5*7/2=7/10
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
a: \(=\dfrac{6}{7}\cdot\dfrac{-3}{5}=\dfrac{-18}{35}\)
b: \(=\dfrac{2}{5}\cdot\dfrac{-15}{8}=\dfrac{-30}{40}=-\dfrac{3}{4}\)
c: \(=\dfrac{2}{4}\cdot\dfrac{7}{3}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)
d: \(=\dfrac{8}{3}\cdot\dfrac{16}{4}=\dfrac{128}{12}=\dfrac{32}{3}\)
theo mình :
a, KQ: -717
b, KQ: -100