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a) 7/10+6/10\(=\frac{13}{10}\)
b) -5/9+2/9\(=\frac{-1}{3}\)
c)1/7+1 6/7\(=2\)
d) 3/13+10/3+1\(=\frac{71}{26}\)
e) 5/12-7/12\(=\frac{-1}{6}\)
f) 7/6+5/6-1/6\(=\frac{11}{6}\)
g) -3/20+7/20-9/20\(=\frac{-1}{4}\)
học tốt
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
a) \(\dfrac{7}{-25}+\dfrac{8}{25}=\dfrac{-7}{25}+\dfrac{8}{25}=\dfrac{1}{25}\)
\(\dfrac{4}{5}+\dfrac{4}{-18}=\dfrac{4}{5+\left(-18\right)}=\dfrac{4}{-13}=\dfrac{-4}{13}\)
\(\dfrac{7}{21}+\dfrac{9}{-36}=\dfrac{1}{3}+\dfrac{1}{-4}=\dfrac{1}{3+\left(-4\right)}=\dfrac{1}{-1}=1\)
b) \(\dfrac{12}{20}-\dfrac{2}{5}=\dfrac{12}{20}+\left(\dfrac{-2}{5}\right)=\dfrac{3}{5}+\dfrac{-2}{5}=\dfrac{3+\left(-2\right)}{5}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-\dfrac{-5}{6}=\dfrac{2}{3}+\dfrac{5}{6}=\dfrac{4}{6}+\dfrac{5}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(6-\dfrac{3}{20}=6+\dfrac{-3}{20}=\dfrac{60}{60}+\dfrac{-9}{60}=\dfrac{51}{60}\)
c) \(\dfrac{2}{21}.\dfrac{-7}{3}=\dfrac{2.\left(-1\right)}{7.3}=\dfrac{-2}{21}\)
\(\dfrac{27}{28}.\left(-21\right)=\dfrac{27.\left(-21\right)}{28}=\dfrac{-567}{28}\)
\(\dfrac{-15}{7}.\dfrac{-14}{25}=\dfrac{-3.\left(-2\right)}{1.5}=\dfrac{6}{5}\)
Bài 1
a: 11/12=1-1/12
23/24=1-1/24
mà -1/12>-1/24
nên 11/12>23/24
b: -3/20=-9/60
-7/12=-35/60
mà -9>-35
nên -3/20>-7/12
a: =>x=7-20=-13
b: =>x=-18+12=-6
c: =>x=9 hoặc x=-6
d: =>x=0 hoặc x=4
e: =>6-x=13-3+14=24
=>x=-18
Câu g và h đề thiếu rồi bạn
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
\(\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)
\(=\left(\frac{157}{8}:\frac{7}{12}-\frac{53}{4}:\frac{7}{12}\right).\frac{4}{5}\)
\(=\left[\left(\frac{157}{8}-\frac{53}{4}\right):\frac{7}{12}\right].\frac{4}{5}\)
\(=\left[\frac{51}{8}:\frac{7}{12}\right].\frac{4}{5}\)
\(=\frac{153}{14}.\frac{4}{5}\)
\(=\frac{306}{35}\)
\(\left(\frac{-2}{5}+\frac{3}{7}\right)-\left(\frac{4}{9}+\frac{12}{20}-\frac{13}{35}\right)+\frac{7}{35}\)
\(=\frac{1}{35}-\frac{212}{315}+\frac{7}{35}\)
\(=\frac{1}{35}+\frac{-212}{315}+\frac{7}{35}\)
\(=\frac{9}{315}+\frac{-212}{315}+\frac{63}{315}\)
\(=\frac{-140}{315}=\frac{-4}{9}\)
B= \(\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+...+\dfrac{7}{1980}\)
B= \(\dfrac{7}{2\times3}+\dfrac{7}{3\times4}+\dfrac{7}{4\times5}+\dfrac{7}{5\times6}+...+\dfrac{7}{44\times45}\)
B= \(7\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{44\times45}\right)\)
B= 7\(\times\) \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{44}-\dfrac{1}{45}\right)\)
B= 7\(\times\) \(\left(\dfrac{1}{2}-\dfrac{1}{45}\right)\)
B= 7\(\times\) \(\dfrac{43}{90}\)
B= \(\dfrac{301}{90}\)