Tìm giá trị của các biểu thức sau
2^7*9^3/6^5 *8^2
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\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7..3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
a) Các phép chia sai: 32 : 6 = 5 (dư 1); 9 : 8 = 1 (dư 0).
Sửa lại:
32 : 6 = 5 (dư 2)
9 : 8 = 1 (dư 1)
b) Ta có thể đặt dấu ngoặc như sau:
(3 + 4) × 9 = 63
9 : (3 + 6) = 1
(16 – 16) : 2 = 0
12 : (3 × 2) = 2
a) \(\frac{2^7}{6^5}\times\frac{9^3}{8^8}=\frac{2^7}{2^5\times3^5}\times\frac{3^6}{2^{24}}=\frac{2^7\times3^6}{2^{29}\times3^5}=\frac{3}{4194304}\)
b) \(\frac{6^3+3\times6^2+3^3}{-13}=\frac{2^3\times3^3+3\times2^2\times3^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{27\times13}{-13}=-27\)
\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot3^6}{3^5\cdot2^5\cdot2^6}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)