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A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)
\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)
\(-5B-B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}-\)\(\left[\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\right]\)
\(-6B=\left(-5\right)^0-\left(-5\right)^{2018}\)
\(B=\left(5^{2018}-1\right):6\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
\(B=\frac{\left[\frac{2}{3}\right]^3\cdot\left[-\frac{3}{4}\right]^2\cdot\left[-1\right]^5}{\left[\frac{2}{5}\right]^2\cdot\left[-\frac{5}{12}\right]^3}\)
\(=\frac{\frac{2^3}{3^3}\cdot\frac{\left[-3\right]^2}{4^2}\cdot\left[-1\right]}{\frac{2^2}{5^2}\cdot\frac{\left[-5\right]^3}{12^3}}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\cdot3\right]^3}}\)
\(=\frac{\frac{1}{3}\cdot\frac{1}{2}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\right]^3\cdot3^3}}\)
\(=\frac{\frac{1\cdot1\cdot\left[-1\right]}{3\cdot2\cdot1}}{\frac{4}{25}\cdot\frac{-125}{4^3\cdot3^3}}\)
\(=\frac{\frac{-1}{6}}{\frac{4}{25}\cdot\frac{-125}{64\cdot27}}=\frac{\frac{-1}{6}}{\frac{4}{1}\cdot\frac{-5}{64\cdot27}}\)
\(=\frac{\frac{-1}{6}}{4\cdot\frac{-5}{64\cdot27}}=\frac{\frac{-1}{6}}{-\frac{20}{64\cdot27}}=\frac{72}{5}\)
\(P=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)
\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)
\(-6B=\left(-5\right)^{2017}-1\)
\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)
Ta có : B = (-5)^0 + (-5)^1 + ......+ (-5)^2017
(-5)B = (-5)^1 + (-5)^2 + .......+ (-5)^2018
(-4)B = (-5)^0- (-5)^2018
B = 1 - (-5)^2018 / (-4)
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