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\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)
=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)
=\(\dfrac{450}{119}\)
Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)
\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)
\(\frac{5^2.3^{11}.2^{11}.2^8+2^2.3^2.3^2.5^2}{2.2^{12}.3^{12}.2^4.5^4-3^8.2^{18}.3^3.5^3}\)
\(=\frac{5^2.3^{11}.2^{19}+2^2.3^4.5^2}{2^{17}.3^{12}.5^4-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.3^4.2^2.\left(3^7.2^{17}+1\right)}{2^{17}.3^{11}.5^3.\left(3.5-2\right)}=\frac{3^7.2^{17}}{2^{15}.3^7.5}+\frac{1}{2^{15}.3^7.5}=\frac{4}{5}+\frac{1}{2^{15}.3^7.5}\)
=5^2.3^11.2^11.2^8.2^2.3^2.3^6.2^12.3^2.5^2/2.3^12.2^12.2^4.5^4-3^8.3^3.2^6.5
ban nhom cac so co cung co so o tu va mau thanh cac nhom va rut gon
Tách phần lử trên ra sao cho có thể rút gọn với phần ơn dưới
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
