Ac. Có bài giải lúc nào vậy.

Ta có   \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\)  \(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)

\(\Leftrightarrow\)  \(c\left(a+b\right)\left(a+b+c\right)+ab\left(a+b\right)=0\)

\(\Leftrightarrow\)  \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\)  a = -b hoặc b = -c hoặc c = -a

1) Nếu a = -b thì  \(a^{2n+1}+b^{2n+1}=-b^{2n+1}+b^{2n+1}=0\)và  \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}=\frac{1}{-b^{2n+1}}+\frac{1}{b^{2n+1}}=0\)

\(\Rightarrow\)  \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}\)

Tương tự cho 2 trường hợp còn lại suy ra đpcm.

ĐK: a,b,c khác 0 và a+b+c khác 0

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\Rightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\Rightarrow\left(a+b\right)\left(\frac{c^2+ca+ab+bc}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(a+b+c\right)}=0\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> hoặc a=-b hoặc b=-c hoặc c=-a.

Khi đó đẳng thức:

\(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}\)đúng với mọi lũy thừa lẻ 2n+1. ĐPCM.

Khó zữ

a)(a+b+c)(ab+bc+ac)-abc=a(ab+bc+ac)+b(ab+bc+ac)+c(ab+bc+ac)-abc

=a²b+abc+a²c+ab²+b²c+abc+abc+bc²+ac²-abc

=(abc+a²b)+(a²c+ac²)+(b²c+ab²)+(bc²+abc)+(abc-abc)

=ab(c+a)+ac(c+a)+b²(c+a)+bc(c+a)

=(ab+ac+b²+bc)(c+a)

=(a+b)(b+c)(c+a)

a) \(\left(a+b+c\right)\left(ab+bc+ac\right)-abc=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+c^2b+c^2a-abc\)

\(=a^2b+ab^2+b^2c+bc^2+c^2a+a^2c+2abc=b\left(a^2+2ac+c^2\right)+b^2\left(a+c\right)+ac\left(a+c\right)\)

\(=b\left(a+c\right)^2+b^2\left(a+c\right)+ac\left(a+c\right)=\left(a+c\right)\left(ab+bc+b^2+ac\right)\)

\(=\left(a+c\right)\left[b\left(a+b\right)+c\left(a+b\right)\right]=\left(a+c\right)\left(a+b\right)\left(b+c\right)\)

b) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)(áp dụng từ câu a) )

\(\Rightarrow a+b=0\)hoặc \(b+c=0\)hoặc \(c+a=0\)

Đặt \(a^{2n+1}=x;b^{2n+1}=y;c^{2n+1}=z\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)( áp dụng câu a) )

\(\Rightarrow x+y=0\)hoặc \(y+z=0\)hoặc \(z+x=0\)

• Với \(x+y=0\Leftrightarrow a^{2n+1}+b^{2n+1}=0\Leftrightarrow\left(a+b\right).A=0\)với A là một đa thức 

Mà ta lại có \(a+b=0\left(cmt\right)\)\(\Rightarrow\)\(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}=0\)\(\Rightarrow\frac{1}{c^{2n+1}}=\frac{1}{c^{2n+1}}\)(luôn đúng)

Tương tự với các trường hợp còn lại, ta có điều phải chứng minh.

\(\)

A >1 là chắc chắn rồi cần gì phải CM nữa cho khổ

thuộc n sao rồi mà

\(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)

\(=\frac{1}{2n}\left[\frac{2n-1+1}{1\left(2n-1\right)}+\frac{2n-3+3}{3\left(2n-3\right)}+...+\frac{3+2n-3}{\left(2n-3\right).3}+\frac{1+2n-1}{\left(2n-1\right).1}\right]\)

\(=\frac{1}{2n}\left(1+\frac{1}{2n-1}+\frac{1}{3}+\frac{1}{2n-3}+...+\frac{1}{2n-3}+\frac{1}{3}+\frac{1}{2n-1}+1\right)\)

\(=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{1}{n}\).

\(M=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-a\right)}\)

Đánh giá đại diện: \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}\)

Tương tự: \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}\)

                   \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)

\(\Rightarrow M=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2N\left(đpcm\right)\)