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mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
A = \(\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)= \(\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
= \(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)= \(\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)= \(\frac{2.6}{3.7}=\frac{4}{7}\)
c, theo đề bài ta có :
x2 = yz, y2 = xz , z2 = xy
\(\Rightarrow\frac{x}{y}=\frac{z}{x},\frac{y}{x}=\frac{z}{y},\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)
AD t/c DTSBN, ta có
\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{X+z+y}{y+x+z}=1\)
x= 1y
z= 1x
y= 1z
=> x = y = x
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
\(x=\frac{\left[6+3^2.2-6.3^2\right]^2}{\left[3^2+3.3^2-3^4\right]^2}=\frac{\left[6\left[1+3-9\right]\right]^2}{\left[9+27-81\right]^2}=\frac{\left[-30\right]^2}{\left[-45\right]^2}=\frac{900}{2025}=\frac{4}{9}\)
a) \(\frac{2^{10}.9^6}{4^6.3^{11}}=\frac{2^{2.5}.3^{2.6}}{4^6.3^{11}}=\frac{\left(2^2\right)^5.3^{12}}{4^6.3^{11}}=\frac{4^5.3^{12}}{4^6.3^{11}}=\frac{3}{4}\)
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^6.3^4}{3^8.\left(3^4\right)^4.3^5.\left(2^3\right)^2}\)
\(M=\frac{3^8.3^{15}.3^6.3^4}{3^{18}.3^{16}.3^5.2^6}=\frac{81}{64}\)
\(N=\frac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(N=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}\)
\(N=\frac{4}{5}\)