Rút gọn
\(A=\dfrac{6^{20}.5^{10}-15^{10}.2^{20}.3^8}{9^{10}.109+109.27^7}\)
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\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)
Ở câu a) số mũ lúc nào cug dương mà bạn ( 45-10 = 4510). Nếu số mũ là dương thì:
a)\(\frac{45^{10}.5^{20}}{75^{15}}\)
= \(\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}\)
= \(\frac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}\)
= \(\frac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
= \(\frac{3^5.1}{1.1}\)
= \(\frac{243}{1}\)
= 243
b)\(\frac{2^{15}.9^4}{6^6.8^2}\)
= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^2}\)
= \(\frac{2^{15}.3^8}{2^6.3^6.2^6}\)
= \(\frac{2^{15}.3^8}{2^{12}.3^6}\)
= \(\frac{2^3.3^2}{1.1}\)
= \(\frac{8.9}{1}\)
= \(\frac{72}{1}\)
= 72
d)
\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)
e)
\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)
f)
\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)
\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)
\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)
\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)
\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)
\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)
\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)
\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)
Đặt A = 1 + 3 + 5 + ... + 97 + 99
Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)
Tổng A bằng: (99 + 1) . 50 : 2 = 2500
Thay A = 2500 vào biểu thức (1), ta được:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
Bài làm:
Ta có: \(\frac{7.6^{10}.2^{20}.3^6-2^{19}.6^{15}}{9.6^{19}.2^9-4.3^{17}.2^{26}}\)
\(=\frac{7.2^{10}.3^{10}.2^{20}.3^6-2^{19}.2^{15}.3^{15}}{3^2.2^{19}.3^{19}.2^9-2^2.3^{17}.2^{26}}\)
\(=\frac{2^{30}.3^{16}.7-2^{34}.3^{15}}{2^{28}.3^{21}-2^{28}.3^{17}}\)
\(=\frac{2^{30}.3^{15}\left(3.7-2^4\right)}{2^{28}.3^{17}\left(3^4-1\right)}\)
\(=\frac{2^2\left(21-16\right)}{3^2\left(81-1\right)}\)
\(=\frac{2^2.5}{3^2.80}\)
\(=\frac{1}{36}\)