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\(\frac{1.3.5+2.6.10+4.12.20}{1.5.7+2.10.14+4.20.28}\)
\(=\frac{3.5+2.3.2.5.2+4.3.4.5.4}{5.7+2.5.2.2.7+4.4.5.7.4}\)
\(=\frac{3.5.\left(1+2.2.2+4.4.4\right)}{5.7.\left(1+2.2.2+4.4.4\right)}\)
\(=\frac{3}{7}>\frac{3}{8}\)
\(A=\frac{\text{1.5.6 + 2.10.12 + 3.15.18 + 4.20.24 + 5.25.30}}{\text{1.3.5 + 2.6.10 + 3.9.15 + 4.12.20 + 5.15.25 }}\)
\(=\frac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
\(=\frac{30.\left(1+2+3+4+5\right)}{15.\left(1+2+3+4+5\right)}\)
\(=\frac{30}{15}=2\)
Vậy A=2.
\(=\frac{1.5.6+\left(1.5.6\right).2+\left(1.5.6\right).3+\left(1.5.6\right).4+\left(1.5.6\right).5}{1.3.5+\left(1.3.5\right).2+\left(1.3.5\right).3+\left(1.3.5\right).4+\left(1.3.5\right).5}\)
\(=\frac{\left(1.5.6\right).\left(1+2+3+4+5\right)}{\left(1.3.5\right).\left(1+2+3+4+5\right)}=\frac{1.5.6}{1.3.5}=\frac{1.1.2}{1.1.1}=2\)
A= \(\frac{1.5.3.2+2.10.2.6+2.15.9.2+4.20.12.2+5.25.15.2}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)
A= \(\frac{2+2+2\cdot2+2+2}{0+0+3+0+0}\)
A= \(\frac{12}{3}\)
A= 4
Đầu tiên bạn tách ra, rút gọn rồi cộng lại,tính nha!
A=\(\dfrac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
A=\(\dfrac{\left(1+2+3+4+5\right).\left(1.5.6\right)}{\left(1+2+3+4+5\right).\left(1.3.5\right)}\) = \(\dfrac{1.5.6}{1.3.5}\) = 2
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(A=\dfrac{15\left(1+2\cdot4+64\right)}{35+240+2240}\)
\(=\dfrac{15\cdot73}{2515}=\dfrac{15\cdot73}{5\cdot503}=\dfrac{3\cdot73}{503}=\dfrac{219}{503}>\dfrac{3}{8}\)