Cho A=1-2+2^2-2^3+...+2^2018
Tinh 3A-2^2019
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Do a là nghiệm của pt nên
\(a^2-a-1=0\Leftrightarrow a^2=a+1\Leftrightarrow a^6=\left(a+1\right)^3=a^3+3a^2+3a+1\)
Và \(a^2-a-1=0\Leftrightarrow a^2-a=1\)
\(P=\dfrac{a^6-3a^3\left(a^2-a\right)-a^3+2018}{a^6-\left(a^3+3a^2+3a+1\right)+2020}=\dfrac{\left(a+1\right)^3-4a^3+2018}{\left(a+1\right)^3-\left(a+1\right)^3+2020}\)
\(P=\dfrac{-3a^3+3a^2+3a+2019}{2020}=\dfrac{-3a\left(a^2-a-1\right)+2019}{2020}=\dfrac{2019}{2020}\)
\(\left\{{}\begin{matrix}a^3+3ab^2=2019\\b^3+3a^2b=2018\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^3+3a^2b+3ab^2+b^3=4037\\a^3-3a^2b+3ab^2-b^3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3=4037\\\left(a-b\right)^3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=\sqrt[3]{4037}\\a-b=1\end{matrix}\right.\Rightarrow a^2-b^2=\left(a-b\right)\left(a+b\right)=\sqrt[3]{4037}\)
Bài 1:
\(A=\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{(a-b)(a+b)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=(2a).\frac{a^2+b^2+a^2-b^2}{(a^2-b^2)(a^2+b^2)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=4a^3.\frac{a^4+b^4+a^4-b^4}{(a^4-b^4)(a^4+b^4)}+\frac{8a^7}{a^8+b^8}=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}=8a^7.\frac{a^8+b^8+a^8-b^8}{(a^8-b^8)(a^8+b^8)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
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\(B=\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}=\frac{(a+1)-a}{a(a+1)}+\frac{(a+2)-(a+1)}{(a+1)(a+2)}+\frac{(a+3)-(a+2)}{(a+2)(a+3)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}\)
\(=\frac{1}{a}-\frac{1}{a+3}=\frac{3}{a(a+3)}\)
Bài 2:
Bạn tham khảo lời giải tương tự tại link sau:
Câu hỏi của Law Trafargal - Toán lớp 8 | Học trực tuyến
\(A=\frac{2^{2019}}{2^{2020}-1}=\frac{1}{2}\left(\frac{2^{2020}-1+1}{2^{2020}-1}\right)=\frac{1}{2}\left(1+\frac{1}{2^{2020}-1}\right)\)
\(B=\frac{3^{2019}}{3^{2020}-1}=\frac{1}{3}\left(1+\frac{1}{3^{2020}-1}\right)< \frac{1}{2}\left(1+\frac{1}{3^{2020}-1}\right)< \frac{1}{2}\left(1+\frac{1}{2^{2020}-1}\right)\)
\(\Rightarrow B< A\)
\(2A=2-2^2+2^3-...+2^{2019}\)
\(\Rightarrow2A+A=3A=2^{2019}+1\)
\(\Rightarrow3A-2^{2019}=2^{2019}+1-2^{2019}=1\)