rút gọn:
\(\frac{8^2.4^5}{2^{20}}\)
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\(\frac{8^2.4^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)
\(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)
\(A=\frac{8^{10}+4^{10}}{8^4+4^{11}}\\ =\frac{8^4.8^6+4^{10}}{8^4+4^{10}.4}\\ =\frac{8^6}{4}\\ =\frac{262144}{4}\\ =65536\)
\(\frac{2^{15}.5^8+2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}=\frac{2^{14}.5^8\left(2+5\right)}{2^{16}.5^7\left(1+5\right)}=\frac{2^{14}.5^8.7}{2^{16}.5^7.6}=\frac{5.7}{4.6}=\frac{35}{24}\)
Ta có: \(-\dfrac{3}{8}x^2z\cdot\dfrac{2}{3}xy^2\cdot z^2\cdot\dfrac{4}{5}x^2y\)
\(=\left(-\dfrac{3}{8}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\right)\cdot\left(x^2\cdot x\cdot x^2\right)\cdot\left(y^2\cdot y\right)\cdot\left(z\cdot z^2\right)\)
\(=-\dfrac{1}{5}x^5y^3z^3\)
Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )
\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)
\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{\left(2^3\right)^5.5^8+\left(-2\right)^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{15}.5^8-2^{14}.5^9}{2^{16}.5^7\left(1+5\right)}\)
\(=\frac{2^{14}.5^8\left(2-5\right)}{2^{16}.5^7.\left(1+5\right)}\)
\(=\frac{2^{14}.5^8.\left(-3\right)}{2^{16}.5^7.6}\)
\(=\frac{-5}{8}\)
\(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)
\(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}\)
\(=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}\)
\(=\frac{1}{2^4}=\frac{1}{16}\)