Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\). Chứng minh \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)với n lẻ.
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28 tháng 11 2018
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\Leftrightarrow a^2b+ab^2+abc+ac^2+abc+ac^2+abc+b^2c+bc^2=abc\Leftrightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc=0\Leftrightarrow\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(b^2c+abc\right)+\left(ac^2+bc^2\right)=0\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
TH1:a=-b
\(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n}-\dfrac{1}{a^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)(vì n lẻ)
\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{a^n-a^n+c^n}=\dfrac{1}{c^n}\)(vì n lẻ)
Suy ra \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)
Chứng minh tương tự trong các trường hợp b=-c và c=-a
Vậy \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)
28 tháng 9 2021
\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)
5 tháng 1 2018
Bài toán tổng quát: Đề này n lẻ mới đúng nhé
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Nếu \(a=-b\Rightarrow a^n=-b^n\) và \(\dfrac{1}{a^n}=\dfrac{-1}{b^n}\)
Ta có: \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)
\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{c^n}\)
VT = VP => ĐPCM
Còn ý còn lại thì dựa trên bài này mà biến đổi một tí là ra
3 tháng 6 2022
2: \(A=9^n\cdot81-9^n+3^n\cdot9+3^n\)
\(=9^n\cdot80+3^n\cdot10\)
\(=10\left(9^n\cdot8+3^n\right)⋮10\)
17 tháng 6 2018
a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
NL
17 tháng 6 2018
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
17 tháng 5 2018
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) hả Lặng Thầm
Lời giải:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow \frac{(a+b)[c(a+b+c)+ab]}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Xét : \(A=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}-\frac{1}{a^n+b^n+c^n}\)
\(A=\frac{a^n+b^n}{a^nb^n}+\frac{a^n+b^n}{c^n(a^n+b^n+c^n)}\)
\(A=(a^n+b^n)\left(\frac{1}{a^nb^n}+\frac{1}{c^n(a^n+b^n+c^n)}\right)\)
\(A=\frac{(a^n+b^n)[c^n(a^n+b^n+c^n)+a^nb^n]}{a^nb^nc^n(a^n+b^n+c^n)}\)
\(A=\frac{(a^n+b^n)(b^n+c^n)(c^n+a^n)}{a^nb^nc^n(a^n+b^n+c^n)}\)
Vì $n$ lẻ nên :
\((a^n+b^n)(b^n+c^n)(c^n+a^n)=(a+b)(b+c)(c+a)(a^{n-1}+....+b^{n-1})(b^{n-1}+..+c^{n-1})(c^{n-1}+...+a^{n-1})\)
\(=0\) do \((a+b)(b+c)(c+a)=0\)
Do đó: \(A=0\Leftrightarrow \frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)