Tìm giá trị các biểu thức sau : \(\frac{4^2.4^3}{2^{10}}\)
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\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}\)
\(=\frac{2^4.2^6}{2^{10}}\)
\(=\frac{2^{10}}{2^{10}}=1\)
hai vế mỗi vế có kết qua bằng 1
khi cộng 2 vầ ta có kết quả chính bằng 2
vậy thôi
dễ
\(\frac{4^2.4^3}{2^{10}}+\frac{3^2.3^3}{3^5}\)
\(=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}+\frac{3^{2+3}}{3^5}\)
\(=\frac{2^4.2^6}{2^{10}}+\frac{3^5}{3^5}\)
\(=\frac{2^{4+6}}{2^{10}}+1\)
\(=\frac{2^{10}}{2^{10}}+1\)
\(=1+1\)
\(=2\)
4^2.4^3/2^10=(2^2)^2/(2^2)^10=2^4/2^20=1/2^16
Nho cho minh nhe
Yen tam minh hoc roi khong sai dau
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
a) \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b) \(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{3^5.0,2^5}{0,2^6}=\frac{3^5}{0,2}=\frac{243}{0,2}=1215\)
c) \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3}{2^{11}}=\frac{3}{2^4}=\frac{3}{16}\)
d) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^2}{-13}=\frac{3^2\left(6^2+1\right)}{-13}=\frac{9.37}{-13}=\frac{333}{-13}\)
a,\(\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{4^5}{4^5}=1\)
\(\frac{\left(0.6\right)^5}{\left(0.2\right)^6}=1215\)
còn lại làm đc mà
\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
a, \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{4+6}}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b,\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}\)
c, \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{3^5.2^{11}}=\frac{3}{2^4}\)
d, \(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(2.3\right)^3+3\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)
\(=\frac{2^3.3^3+3^3.2^2+3^3}{-13}=\frac{3^9\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3=27\)
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
\(\frac{4^2.4^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
\(\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{\left(2^2\right)^5}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)