\(\left(\dfrac{1}{3}\right)^{2017}.3^{2016}.21=......\)
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Những câu hỏi liên quan
21 tháng 6 2017
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
22 tháng 6 2017
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
31 tháng 1 2018
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)
\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)
\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)
AH
Akai Haruma
Giáo viên
17 tháng 10 2018
Lời giải:
Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)
\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)
Do đó:
\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)
\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)
............
\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)
Cộng theo vế:
\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)
\(=\underbrace{1+1+1...+1}_{1008}=1008\)
30 tháng 6 2022
\(\Leftrightarrow1-11< =3m< =\left(9-9\right)\cdot A=0\)
=>-10<=3m<=0
hay \(m\in\left\{-3;-2;-1;0\right\}\)
24 tháng 7 2017
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1-\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1}{2017}\)
H
24 tháng 7 2017
Giải:
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right).\left(1-\dfrac{1}{2017}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)
\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)
\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)
Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:
\(A=\dfrac{1}{2017}\)
Vậy \(A=\dfrac{1}{2017}\).
Chúc bạn học tốt!
\(\left(\dfrac{1}{3}\right)^{2017}\cdot3^{2016}\cdot21=\dfrac{1}{3^{2017}}\cdot3^{2016}\cdot21=\dfrac{3^{2016}}{3^{2017}}\cdot21=\dfrac{1}{3}\cdot21=7\)