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d, \(\dfrac{-31}{240}\) e , \(\dfrac{43}{72}\) g, \(\dfrac{-5}{36}\)
\(a,\frac{a}{-b}=\frac{-a}{b}vìa.b=-a.\left(-b\right)\)
\(b,\frac{-a}{-b}=\frac{a}{b}vì-a.b=-b.a\)
a , \(\frac{3.5}{8.24}=\frac{3.5}{8.8.3}=\frac{5}{64}\)
b , \(\frac{2.14}{7.8}=\frac{2.2.7}{7.2.2.2}=\frac{1}{2}\)
c, \(\frac{3.7.11}{22.9}=\frac{3.7.11}{2.11.3}=\frac{7}{6}\)
d , \(\frac{8.5-8.2}{16}=\frac{8.\left(5-2\right)}{8.2}=\frac{8.3}{8.2}=\frac{3}{2}\)
e , \(\frac{11.4-11}{2-13}=\frac{11.\left(4-1\right)}{-11}=\frac{11.3}{-11}=-3\)
C=c.3/4+c.5/6-c.19/12
=c.(3/4+5/6-19/12)
=c.0
=0
Vậy C=0
Thay c = 2002/2003 vào, ta có:
C = 2002/2003 . 3/4 + 2002/2003 . 5/6 - 2002/2003 . 19/12
C = 2002/2003 . (3/4 + 5/6 - 19/12)
C = 2002/2003 . (9/12 + 10/12 - 19/12)
C = 2002/2003 . 0
C = 0
\(\dfrac{-5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-7}{21}+\dfrac{8}{24}=\dfrac{-1}{3}+\dfrac{1}{3}=\dfrac{0}{3}=0\)\(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}=(\dfrac{-3}{7}+\dfrac{-4}{7})+\dfrac{5}{13}=\dfrac{-7}{7}+\dfrac{5}{13}=\left(-1\right)+\dfrac{5}{13}=\dfrac{-13}{13}+\dfrac{5}{13}=\dfrac{-8}{13}\)
a) \(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\dfrac{5}{13}\)
= -1+\(\dfrac{5}{13}\)
=\(\dfrac{-13}{13}+\dfrac{3}{13}\)
=\(\dfrac{-8}{13}\)
a) \(\frac{3\cdot5}{8\cdot24}=\frac{5}{8\cdot8}=\frac{5}{64}\)
b) \(\frac{2\cdot14}{7\cdot8}=\frac{14}{7\cdot4}=\frac{2}{4}=\frac{1}{2}\)
c) \(\frac{3\cdot7\cdot11}{22\cdot9}=\frac{7}{2\cdot3}=\frac{7}{6}\)
d) \(\frac{8\cdot5-8\cdot2}{16}=\frac{8\cdot3}{16}=\frac{24}{16}=\frac{3}{2}\)
e) \(\frac{11\cdot4-11}{2-13}=\frac{11\cdot3}{2-13}=\frac{33}{-11}=-3\)
\(\frac{3}{-4}=\frac{-3}{4};\frac{-5}{-7}=\frac{5}{7};\frac{2}{-9}=\frac{-2}{9};\frac{-11}{-10}=\frac{11}{10}\)
\(a.\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\left(\frac{1}{3}x\right):\frac{2}{3}=\frac{35}{8}\)
\(\left(\frac{1}{3}x\right)=\frac{35}{8}:\frac{2}{3}\)
\(\left(\frac{1}{3}x\right)=\frac{35}{12}\)
\(x=\frac{35}{12}:\frac{1}{3}\)
\(x=\frac{35}{4}\)
\(b.4,5:0,3=2,25:\left(0,1.x\right)\)
\(\frac{9}{2}:\frac{3}{10}=\frac{9}{4}:\left(\frac{1}{10}x\right)\)
\(15=\frac{9}{4}:\left(\frac{1}{10}x\right)\)
\(\frac{1}{10}x=\frac{9}{4}:15\)
\(\frac{1}{10}x=\frac{3}{20}\)
\(x=\frac{3}{20}:\frac{1}{10}\)
\(x=\frac{3}{2}\)
\(c.8:\left(\frac{1}{4}x\right)=2:0,02\)
\(8:\left(\frac{1}{4}x\right)=2:\frac{2}{100}\)
\(8:\left(\frac{1}{4}x\right)=100\)
\(\frac{1}{4}x=8:100\)
\(\frac{1}{4}x=\frac{2}{25}\)
\(x=\frac{2}{25}:\frac{1}{4}\)
\(x=\frac{8}{25}\)
\(d.3:2\frac{1}{4}=\frac{3}{4}:\left(6.x\right)\)
\(\frac{27}{4}=\frac{3}{4}:\left(6.x\right)\)
\(6x=\frac{3}{4}:\frac{27}{4}\)
\(6x=\frac{1}{9}\)
\(x=\frac{1}{9}:6\)
\(x=\frac{1}{54}\)
a)
b) 4,5 : 0,3 = 2,25 : ( 0,1.x) => 0,1.x =
c)
d)
Chúc bạn học tốt!
Giải:
a)
b)
;
c)
.