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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\Leftrightarrow c\left(a+b+c\right)\left(a+b\right)=-ab\left(a+b\right)\)
\(\Leftrightarrow\left(ac+bc+c^2\right)\left(a+b\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> a=-b hoặc b=-c hoặc c=-a
không mất tính tổng quát ,giả sử a=-b, ta có:
\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{-b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\left(1\right)\)
\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\left(2\right)\)
Từ (1) và (2) => đpcm
Tương tự với 2 trường hợp còn lại ta cũng có đpcm
\(\frac{1}{a}+\frac{1}{c}=\frac{1}{a-b+c}+\frac{1}{b}\Leftrightarrow\frac{a+c}{ac}=\frac{a+c}{b\left(a-b+c\right)}\)
\(\Rightarrow\left[{}\begin{matrix}a+c=0\\ac=b\left(a-b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-c\\ac=b\left(a-b\right)+bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\ac-bc-b\left(a-b\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-c\\\left(c-b\right)\left(a-b\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\a=b\left(l\right)\\b=c\left(l\right)\end{matrix}\right.\) do \(a< b< c\) \(\Rightarrow a=-c\)
\(\Rightarrow\frac{1}{a^{2019}}-\frac{1}{b}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}}-\frac{1}{b}-\frac{1}{a^{2019}}=\frac{-1}{b}\)
\(\frac{1}{a^{2019}-b+c^{2019}}=\frac{1}{a^{2019}-b-c^{2019}}=\frac{-1}{b}\)
\(\Rightarrow\frac{1}{a^{2019}}-\frac{1}{b}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}-b+c^{2019}}\)
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
Méo bt trẩu là gì à =))
Bảo ezzz thì chỉ hộ cách làm ko bt thì đừng cư xử như 1 đứa trẻ trâu=))
Câu hỏi của hanhungquan - Toán lớp 8 - Học toán với OnlineMath tương tự
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2019}\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{2019}\Leftrightarrow2019\left(ab+bc+ca\right)=abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)
\(\Leftrightarrow\left(ab+bc\right)\left(a+b+c\right)+ca\left(a+b+c\right)-abc=0\)
\(\Leftrightarrow b\left(a+c\right)\left(a+b+c\right)+ca\left(a+c\right)+abc-abc=0\)
\(\Leftrightarrow\left(a+c\right)\left(ab+b^2+bc+ca\right)=0\)
\(\Leftrightarrow\left(a+c\right)\left[b\left(a+b\right)+c\left(a+b\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a+b=0\)hoặc \(b+c=0\)hoặc \(c+a=0\)
Mà \(a+b+c=2019\)
\(\Rightarrow a=2019\)hoặc \(b=2019\)hoặc \(c=2019\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2019}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{ab+ac+bc+c^2}{abc\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=0\Rightarrow c=2019\\b+c=0\Rightarrow a=2019\\a+c=0\Rightarrow b=2019\end{matrix}\right.\)
Đặt : \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=P\)
\(\Rightarrow\left(a+b+c\right).P=\frac{1}{2019}.2019\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{6057}{2019}+\frac{\left(-4038\right)}{2019}\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=3+\left(-2\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)
\(a+b=c+\frac{1}{2019}\Leftrightarrow a+b-c=\frac{1}{2019}\Leftrightarrow\frac{1}{a+b-c}=2019\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=2019\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{a+b-c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b-c}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{a+b}{c\left(a+b-c\right)}\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)=\left(a+b\right)ab\)
\(\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ca+bc-c^2-ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a-c\right)-b\left(a-c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-b\right)\left(a-c\right)=0\)
=>a=-b hoặc c=b hoặc a=c
không mất tính tổng quát, giả sử a=-b, ta có:
\(P=\left(-b^{2019}+b^{2019}-c^{2019}\right)\left(-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)=\left(-c\right)^{2019}\cdot\left(\frac{-1}{c}\right)^{2019}=1\)
tương tư với các trường hợp khác ta cũng có P=1
Vậy P=1