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c) gọi biểu thức là S = 2 + 2\(^2+2^3+.....+2^{50}\)
2S=2\(^2+2^3+2^4+......+2^{50}+2^{51}\)
\(2S-S=S=2^{51}-2\)
b) \(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{10}}\)
= \(2+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^9}\)
2S-S=S=(\(2+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^9}\))-( \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\))
bạn tự tìm S nhé
mink làm được như thế đó, phần a mink không muốn nhấn mỏi tay bạn ạ, đừng nghĩ mink ko biết làm nha
\(1,\)
\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}\)
\(=\dfrac{3^4.5^2.3^8.2^5.5^5}{5^5.3^7.2^5.3^{10}}\)
\(=\dfrac{3^{12}.2^5.5^7}{5^5.3^{17}.2^5}\)
\(=\dfrac{1.5^2}{3^5.1}\)
\(=\dfrac{25}{243}\)
\(2,\)
\(\dfrac{4^5.9^4+2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8+2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8+2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8.4}{2^{10}.3^8.6}\)
\(=\dfrac{2^{12}.3^8}{2^{11}.3^9}\)
\(=\dfrac{2}{3}\)
\(3,\)
\(\dfrac{15.3^{11}+4.27^4}{9^7}\)
\(=\dfrac{3.5.3^{11}+2^2.3^{12}}{3^{14}}\)
\(=\dfrac{5.3^{12}+2^2.3^{12}}{3^{14}}\)
\(=\dfrac{3^{12}\left(5+2^2\right)}{3^{14}}\)
\(=\dfrac{3^{12}.9}{3^{14}}\)
\(=\dfrac{3^{14}}{3^{14}}\)
\(=1\)
\(4,\)
\(\dfrac{4^7.2^8}{3.2^{15}.16^2-5^2\left(2^{10}\right)^2}\)
\(=\dfrac{2^{22}}{3.2^{23}-5^2.2^{20}}\)
\(=\dfrac{2^{22}}{2^{20}.\left(-1\right)}\)
\(=\dfrac{2^{22}}{-2^{20}}\)
\(=-4\)
* Mấy bài còn lại tương tự đấy bạn tự làm đi
Mình mỏi tay lắm rồi
P/s:khuyến khích tự làm,chỉ làm mẫu 1 câu:
1)\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}=\dfrac{\left(5.9\right)^2.3.3^7.\left(2.5\right)^5}{5^5.3^7.\left(2.9\right)^5}\)\(=\dfrac{5^2.9^2.3.3^7.2^5.5^5}{5^5.3^7.2^5.9^5}\)\(=\dfrac{5^2.9^2.3.1.1.1}{1.1.1.9^5}\)\(=\dfrac{5^2.9^2.3}{9^5}=\dfrac{5^2.9^2.3}{9^2.9^3}=\dfrac{5^2.3}{9^3}=\dfrac{75}{729}=\dfrac{25}{243}\)
a) \(2^{3x+2}=4^{x+5}\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\Leftrightarrow2^{3x+2}=2^{2x+10}\)
\(\Rightarrow3x+2=2x+10\Leftrightarrow3x+2-2x-10\)
\(\Leftrightarrow x-8=0\Leftrightarrow x=8\) vậy \(x=8\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))
\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
\(M=\dfrac{-3}{4}\cdot\dfrac{2}{11}+\dfrac{-3}{4}\cdot\dfrac{9}{11}+2\dfrac{3}{4}\)
\(=\dfrac{-3}{4}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{11}{4}\\ =\dfrac{-3}{4}\cdot1+\dfrac{11}{4}\\ =-\dfrac{3}{4}+\dfrac{11}{4}\\ =\dfrac{8}{4}\\ =2\)
\(N=\dfrac{6}{8}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-4\right)^2\)
\(=\dfrac{6}{8}+\dfrac{1}{8}-\dfrac{3}{16}\cdot16\\ =\dfrac{7}{8}-3\\ =\dfrac{-17}{8}\)
Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)
\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\)
⇒5A=15+252+...+11511⇒5A=15+252+...+11511
⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512
⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512
⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511
⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)
⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1
⇒A<116⇒A<116
\(M=\dfrac{-3}{4}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+2+\dfrac{3}{4}=2\)
\(N=\dfrac{6}{8}+\dfrac{1}{8}-\dfrac{3}{16}\cdot16=\dfrac{7}{8}-3=-\dfrac{17}{8}\)