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d, \(\dfrac{-31}{240}\) e , \(\dfrac{43}{72}\) g, \(\dfrac{-5}{36}\)
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 , \(\frac{11.4-11}{2-13}=\frac{11.\left(4-1\right)}{-11}=\frac{11.3}{-11}=-3\)
\(\dfrac{-5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-7}{21}+\dfrac{8}{24}=\dfrac{-1}{3}+\dfrac{1}{3}=\dfrac{0}{3}=0\)\(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}=(\dfrac{-3}{7}+\dfrac{-4}{7})+\dfrac{5}{13}=\dfrac{-7}{7}+\dfrac{5}{13}=\left(-1\right)+\dfrac{5}{13}=\dfrac{-13}{13}+\dfrac{5}{13}=\dfrac{-8}{13}\)
a) \(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\dfrac{5}{13}\)
= -1+\(\dfrac{5}{13}\)
=\(\dfrac{-13}{13}+\dfrac{3}{13}\)
=\(\dfrac{-8}{13}\)
\(\frac{3}{-4}=\frac{-3}{4};\frac{-5}{-7}=\frac{5}{7};\frac{2}{-9}=\frac{-2}{9};\frac{-11}{-10}=\frac{11}{10}\)
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
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) \(\frac{11\cdot4-11}{2-13}=\frac{11\cdot3}{2-13}=\frac{33}{-11}=-3\)
Hình 26 cho biết hai góc kề bù xOy và yOy', . Tính
Giải:
Hai góc xOy và yOy' kề bù nên
suy ra:
Có: \(\widehat{xOy}+\widehat{yOy'}=180^o\)
=> \(\widehat{yOy'}=180-\widehat{xOy}=180-120=60\)
\(a,\frac{a}{-b}=\frac{-a}{b}vìa.b=-a.\left(-b\right)\)
\(b,\frac{-a}{-b}=\frac{a}{b}vì-a.b=-b.a\)