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a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
\(F=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(2^2-1\right)\)
\(F=1+2+3+4+...+20\)
\(F=21.10=210\)
\(F=\left(20^2+18^2+...+4^2+2^2\right)-\left(19^2+17^2+...+3^2-1\right)\)
\(F=20^2+18^2+....+4^2+2^2-19^2-17^2-....-3^2-1^2\)
\(F=\left(20^2-19^2\right)+\left(18^2-17^2\right)+.....+\left(4^2-3^2\right)+\left(2^2-1^2\right)\)
Áp dụng hằng đẳng thức : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\) ta được:
\(F=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(F=1.39+1.35+....+1.7+1.3=39+35+.....+7+3\)
Dãy trên có: (39-3):4+1=10 (số hạng)
=>\(F=\frac{\left(39+3\right).10}{2}=\frac{420}{2}=210\)
\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)
\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)
\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)
\(=\frac{3^{61}}{3^{60}}\)
\(=3\)
\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)
\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)
\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)
\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)
\(=\frac{3^4}{3^3}\)
\(=3\)
Gọi dãy số trên là : A
\(A=2^{17}-2^{16}-2^{15}-......-2^2-2-1\)
\(\Rightarrow2A=2^{18}-2^{16}-2^{15}-.......-2^2-2\)
\(\Rightarrow A=2A-A=2^{18}-1\)