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Bài 1:
\(\frac{-3}{4}=\frac{\left(-3\right)\cdot5}{4\cdot5}=\frac{-15}{20}\)
\(\frac{4}{-5}=\frac{-4}{5}=\frac{\left(-4\right)\cdot4}{5\cdot4}=\frac{-16}{20}\)
Ta thấy:\(\frac{-15}{20}>\frac{-16}{20}\Leftrightarrow-\frac{3}{4}>-\frac{4}{5}\)
C = \(\frac{1010}{1008.8-994}\)
= \(\frac{1010}{1008.7+1008-994}\)
= \(\frac{1010}{1008.7+14}\)
= \(\frac{1010}{1008.7+2.7}\)
= \(\frac{1010}{7\left(1008+2\right)}=\frac{1010}{7.1010}=\frac{1}{7}\)
D = \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.6+2.6.12+3.9.18+5.15.30}\)
= \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.2.3+2.6.4.3+3.9.6.3+5.15.10.3}\)
= \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{3\left(1.2.3+2.4.6+3.6.9+5.10.15\right)}=\frac{1}{3}\)
BCNN(7,3) = 21
\(\frac{1}{7}=\frac{1.3}{7.3}=\frac{3}{21}\)
\(\frac{1}{3}=\frac{1.7}{3.7}=\frac{7}{21}\)
\(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.6+2.6.12+3.9.18+5.15.30}\)
\(=\frac{1.2.3.\left(1+2+3+5\right)}{1.3.6.\left(1+2+3+5\right)}\)
\(=\frac{1.2.3.5}{1.3.6.5}\)
\(=\frac{1}{3}\)
C = \(\dfrac{1010}{1008\times8-994}\)
C = \(\dfrac{1010}{8064-994}\)
C = \(\dfrac{1010}{7070}\)
C = \(\dfrac{1}{7}\)
D = \(\dfrac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.6+2.6.12+3.9.18+5.15.30}\)
D = \(\dfrac{6+8.6+6.27+6.125}{18+18.8+18.27+18.125}\)
D = \(\dfrac{6.\left(1+8+27+125\right)}{18.\left(1+8+27+125\right)}\)
D = \(\dfrac{1}{3}\)
\(\dfrac{-22}{36}=-\dfrac{11}{18}\)
\(\dfrac{-51}{34}=-\dfrac{3}{2}\)
\(\dfrac{132}{-144}=-\dfrac{11}{12}\)
\(\dfrac{-126}{270}=-\dfrac{7}{15}\)
\(\dfrac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.6+2.6.12+3.9.18+5.15.30}=\dfrac{1.2.3\left(1+2^3+3^3+5^3\right)}{1.3.6\left(1+2^3+3^3+5^3\right)}=\dfrac{1}{3}\)
\(\dfrac{3469-54}{6938-108}=\dfrac{3469-54}{2.\left(3469-54\right)}=\dfrac{1}{2}\)
\(\dfrac{2468-98}{3702-147}=\dfrac{2.\left(1234-49\right)}{3.\left(1234-49\right)}=\dfrac{2}{3}\)