1/a+1+1/a(a+1)

=a(a+1)+(a+1)/(a+1)*a(a+1)

=(a+1)*(a+1)/(a+1)*a(a+1)

=1/a

Xét VP ta có :

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

=> đpcm

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

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

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

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

\(\Rightarrow a+1-a=1\)

\(\Rightarrow1=1\left(đpcm\right)\)

Giữa ac bc là dấu gì vậy bạn?

cần nữa ko nhok . anh giải cho , dễ mà

\(ĐKXĐ:\hept{\begin{cases}a\inℤ\\a\ne0\\a\ne-1\end{cases}}\)

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

\(\Leftrightarrow\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}-\frac{1}{a}=0\)

\(\Leftrightarrow\frac{a+1-a-1}{a\left(a+1\right)}=0\)

\(\Leftrightarrow0=0\left(ĐPCM\right)\)