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\(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{8}=-92\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
\(\frac{3^{10}+6^2}{5\cdot3^8+20}\)
\(=\frac{3^{10}+\left(3 \cdot2\right)^2}{5\cdot3^8+5\cdot4}\)
\(=\frac{3^{10}+3^2\cdot2^2}{5\cdot\left(3^8+4\right)}\)
\(=\frac{3^2\cdot\left(3^8+2^2\right)}{5\cdot\left(3^8+4\right)}\)
\(=\frac{3^2\cdot\left(3^8+4\right)}{5\cdot\left(3^8+4\right)}\)
\(=\frac{3^2}{5}\)
\(=\frac{9}{5}\)
A (-6.37*0.4)*2.5=-6.37*(0.4*2.5)=-6.37*1=-6.37
B (-0.125*-5.3)*8=(-0.125*8)*(-5.3)=-1*(-5.3)=5.3
C -2.5*(-4)*(-7.9)=10*(-7.9)=-79
\(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^2+6^2}{5+20}=\frac{9^2}{25}=\frac{81}{25}=\frac{9}{5}\)
\(\frac{8}{3}\cdot\frac{2}{5}\cdot\frac{3}{8}\cdot10\cdot\frac{19}{92}=\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot\left(\frac{2}{5}\cdot10\right)\cdot\frac{19}{92}=1\cdot4\cdot\frac{19}{92}=4\cdot\frac{19}{92}=\frac{19}{23}\)
\(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}\)
\(=(\frac{8}{3}.\frac{3}{8}).(\frac{2}{5}.10).\frac{19}{92}\)
\(=4.\frac{19}{92}\)
\(=\frac{19}{23}\)
Hok tôt !!!!!!!!!!!!!!!