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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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}\left(đpcm\right)\)
\(\frac{1}{a}=\frac{1}{a+1}+\left(\frac{1}{a+\left(a+1\right)}\right)\)
\(\Rightarrow\frac{1}{a}-\frac{1}{a+1}=\frac{1}{a\left(a+1\right)}\)
\(\Rightarrow\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)
\(\Rightarrow a+1-a=1\)
\(\Rightarrow1=1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}a\inℤ\\a\ne0\\a\ne-1\end{cases}}\)
\(\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)
\(\Leftrightarrow\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}-\frac{1}{a}=0\)
\(\Leftrightarrow\frac{a+1-a-1}{a\left(a+1\right)}=0\)
\(\Leftrightarrow0=0\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Do \(\frac{a}{b}< 1\Leftrightarrow\)a<b \(\Leftrightarrow\)a+n<b+n
Ta có: \(\frac{a}{b}\)= 1 - \(\frac{a-b}{b}\)
\(\frac{a+n}{b+n}\)= 1- \(\frac{a-b}{b+n}\)
Do \(\frac{a-b}{b}\)>\(\frac{a-b}{b+n}\)=> \(\frac{a}{b}\)<\(\frac{a+n}{b+n}\)
2.Tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}=y\)
Đúng 100%
![](https://rs.olm.vn/images/avt/0.png?1311)
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