Cho \(B=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2020}\)
Chứng minh B ko là số nguyên.
Những câu hỏi liên quan
Cho \(A=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2018}\)
Chứng minh A không là số nguyên.
Cho b=\(1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2020}\)
Cmr b<0
Thực hiện phép tính :
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
Thực hiện phép tính
a,\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(2018\cdot\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019\cdot\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-4038-\frac{2019}{2017}+4038\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
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Cho 3 số x,y,z thỏa mãn \(\hept{\begin{cases}x+y+z=2019\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\end{cases}}\).Tính giá trị biểu thức \(P=\left(x^{2017}+y^{2017}\right)\left(y^{2019}+z^{2019}\right)\left(z^{2021}+x^{2021}\right)\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z}{\left(x+y+z\right).z}-\frac{x+y+z}{z.\left(x+y+z\right)}=\frac{-x-y}{z.\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{x+y}{-z.\left(x+y+z\right)}\)
TH1: x+y=0
=> x=-y => P=0
TH2: xy=-z.(x+y+z)
\(\Leftrightarrow xy=-xz-zy-z^2\Leftrightarrow xy+xz+zy+z^2=0\Leftrightarrow x.\left(y+z\right)+z.\left(y+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(y+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-z\\y=-z\end{cases}\Rightarrow P=0}\)
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Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
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\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Leftrightarrow\)\(x+1=2019\)
\(\Leftrightarrow\)\(x=2019-1\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
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Xem thêm câu trả lời
Cho \(A=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\). Chứng tỏ A ko phải là 1 số nguyên.
Mk cần gấp. Mai nộp rồi!!!
sao ko có ai giúp mk vậy
cho a,b,c thỏa mãn: \(\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\)
Tính giá trị biểu thức : A=\(A=\frac{a^{2017}+b^{2018}+c^{2019}}{a^{2017}\times b^{2018}\times c^{2019}}\)
Cho \(S_n=\frac{2}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)Chứng minh rằng: \(S_1+S_2+...+S_{2017}< \frac{2017}{2019}\)