Đây:

Ta có: \(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)

\(=\frac{a+1}{a\left(a+1\right)}\)

\(=\frac{1}{a}\)

Vậy \(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)=\frac{1}{a}}\)

bài này bạn nhân hết ra, hình như phân tích ra mẫu sẽ là\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)và sẽ bằng 0

Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\)

\(\Leftrightarrow ca+cb=2ab\)

\(\Leftrightarrow ac-ab=ab-bc\)

\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)

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

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{c}=\frac{1}{2}\left(\frac{b+a}{2ab}\right)\)

\(\frac{1}{c}=\frac{b+a}{2ab}\)

suy ra \(2ab=c\left(b+a\right)\)

\(2ab=cb+ca\)

suy ra \(ab+ab=cb+ca\)

suy a \(ab-cb=ca-ab\)

suy ra \(b\left(a-c\right)=a\left(c-b\right)\)

suy ra \(\frac{a}{b}=\frac{a-c}{c-b}\left(Đpcm\right)\)

chua hieu may

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{c}=\frac{1}{2}\left(\frac{b+a}{2ab}\right)\)

\(\frac{1}{c}=\frac{b+a}{2ab}\)

\(\Rightarrow2ab=c\left(b+a\right)\)

\(2ab=cb+ca\)

\(\Rightarrow ab+ab=cb+ca\)

\(\Rightarrow ab-cd=ca-ab\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

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

trong phân số mẫu luôn thuộc Z và lớn hơn 0

nên a ∈ Z và a ≠ 0

\(\frac{1}{a+1}\)nếu a=-1 thì \(\frac{1}{-1+1}\)=\(\frac{1}{0}\)mẫu khác 0 nên a ≠ -1

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2\frac{1}{ab}+2\frac{1}{bc}+2\frac{1}{ac}\)

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)

\(\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=0\\ 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\)

\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=0\\ \frac{abc^2+a^2bc+ab^2c}{a^2b^2c^2}=0\)

\(abc^2+a^2bc+ab^2c=0\\ abc\left(c+a+b\right)=0\)

\(a+b+c=0\)(DPCM)

đặt x=a-b;y=b-c;z=c-a

ta có x+y+z=0

nên ta có ĐPCM 

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

<=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

<=> \(2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=0\)

<=> \(\frac{z}{xyz}+\frac{y}{xyz}+\frac{x}{xyz}=0\)

<=> \(\frac{x+y+z}{xyz}=0\) (luôn đúng )