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Giải:
\(\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left(\dfrac{1}{4}\right)^6.2017}{\dfrac{1}{4096}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left[\left(\dfrac{1}{2}\right)^2\right]^6.2017}{\left(\dfrac{1}{2}\right)^{12}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left(\dfrac{1}{2}\right)^{12}.2017}{\left(\dfrac{1}{2}\right)^{12}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.\left[2018-\left(\dfrac{1}{2}\right)^{10}.2017\right]}{\left(\dfrac{1}{2}\right)^{12}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)
\(=\dfrac{2018-\left(\dfrac{1}{2}\right)^{10}.2017}{\left(\dfrac{1}{2}\right)^{10}.\left(-\dfrac{1}{6}\right)}\)
\(=\dfrac{2018}{\left(\dfrac{1}{2}^{10}\right).\left(-\dfrac{1}{6}\right)}-\dfrac{\left(\dfrac{1}{2}\right)^{10}.2017}{\left(\dfrac{1}{2}\right)^{10}.\left(-\dfrac{1}{6}\right)}\)
\(=\dfrac{2018}{\left(\dfrac{1}{2}^{10}\right).\left(-\dfrac{1}{6}\right)}+\dfrac{2017}{\dfrac{1}{6}}\)
\(=-12398592+12102\)
\(=-12386490\)
Vậy ...
Chúc bạn học tốt!
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
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Câu hỏi của Kurosaki Akatsu - Toán lớp 8 - Học toán với OnlineMath
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x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)
\(=\frac{7}{12}\)
b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
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a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
\(\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{4}\right)^6.2017\)
\(=\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{2}\right)^{12}.2017\)
\(=\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{2}\right)^2.\left(\frac{1}{2}\right)^{10}.2017\)
\(=\left(\frac{1}{2}\right)^2.\left[\left(\frac{1}{2}\right)^{10}.\left(2018-2017\right)\right]\)
\(=\left(\frac{1}{2}\right)^2.\left(\frac{1}{2}\right)^{10}\)
\(=\left(\frac{1}{2}\right)^{12}=\frac{1}{2^{12}}\)