ĐKXĐ bạn tự xét nhé

\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)

\(M=\left(\frac{a^2+1}{a^2+1}+\frac{a}{a^2+1}\right):\left(\frac{a^2+1}{\left(a^2+1\right)\left(a-1\right)}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right)\)

\(M=\left(\frac{a^2+a+1}{a^2+1}\right):\left(\frac{a^2+1}{\left(a^2+1\right)\left(a-1\right)}-\frac{2a}{\left(a^2+1\right)\left(a-1\right)}\right)\)

\(M=\left(\frac{a^2+a+1}{a^2+1}\right):\left(\frac{a^2-2a+1}{\left(a^2+1\right)\left(a-1\right)}\right)\)

\(M=\left(\frac{a^2+a+1}{a^2+1}\right):\left(\frac{\left(a-1\right)^2}{\left(a^2+1\right)\left(a-1\right)}\right)\)

\(M=\frac{\left(a^2+a+1\right)\left(a^2+1\right)\left(a-1\right)}{\left(a^2+1\right)\left(a-1\right)^2}\)

\(M=\frac{a^2+a+1}{a-1}\)

Để M thuộc Z thì \(a^2+a+1⋮a-1\)

\(\Leftrightarrow a^2-a+2a-2+3⋮a-1\)

\(\Leftrightarrow a\left(a-1\right)+2\left(a-1\right)+3⋮a-1\)

\(\Leftrightarrow\left(a-1\right)\left(a+2\right)+3⋮a-1\)

Mà \(\left(a-1\right)\left(a+2\right)⋮a-1\)

\(\Rightarrow3⋮a-1\)

\(\Rightarrow a-1\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

\(\Rightarrow a\in\left\{2;4;0;-2\right\}\)

Để M = 7 thì :

\(\frac{a^2+a+1}{a-1}=7\)

\(\Leftrightarrow a^2+a+1=7\left(a-1\right)\)

\(\Leftrightarrow a^2+a+1=7a-7\)

\(\Leftrightarrow a^2-6a+8=0\)

\(\Leftrightarrow a^2-2a-4a+8=0\)

\(\Leftrightarrow a\left(a-2\right)-4\left(a-2\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a-2=0\\a-4=0\end{cases}\Rightarrow\orbr{\begin{cases}a=2\\a=4\end{cases}}}\)

Để M > 0 thì :

\(\frac{a^2+a+1}{a-1}>0\)

Vì \(a^2+a+1>0\forall a\), do đó để M > 0 thì : \(a-1>0\Leftrightarrow a>1\)

Chứng minh \(a^2+a+1>0\):

Đặt \(B=a^2+a+1\)

\(B=a^2+2\cdot a\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)

\(B=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(a+\frac{1}{2}\right)^2\ge0\forall a\)

\(\Rightarrow B\ge0+\frac{3}{4}=\frac{3}{4}>0\)

\(\Rightarrow B>0\left(đpcm\right)\)

Dấu "=" xảy ra \(\Leftrightarrow a+\frac{1}{2}=0\Leftrightarrow a=\frac{-1}{2}\)

Lời giải:

Ta có: \(A=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\)

\(\Leftrightarrow A=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)(a+b+c)\geq (1+1+2)^2\)

\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\geq \frac{4^2}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)

Do đó: \(A\geq 3+\frac{8}{3}=\frac{17}{3}\) hay \(A_{\min}=\frac{17}{3}\)

Dấu bằng xảy ra khi \((a,b,c)=(\frac{3}{2}; \frac{3}{2}; 3)\)

\(P=a^2+a+1\)

\(=a^2+\frac{1}{2}\cdot2\cdot a+\frac{1}{4}+\frac{3}{4}\)

\(=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\left(a+\frac{1}{2}\right)^2\ge0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow P\ge\frac{3}{4}\)

dấu "=" xảy ra khi : 

\(\left(a+\frac{1}{2}\right)^2=0\Rightarrow a+\frac{1}{2}=0\Rightarrow a=-\frac{1}{2}\)

vậy 

cho A=6n-1/3n+1(n thuoc z) hoi a tim n de A nguyen b tim n de A co gia tri nho nhat

Giải:Ta có:A=\(\frac{6n-1}{3n+1}=\frac{6n+2-3}{3n+1}=\frac{2\left(3n+1\right)}{3n+1}-\frac{3}{n+1}=2-\frac{3}{n+1}\)

a,Để A nguyên thì \(\frac{3}{n+1}\in Z\)\(\Rightarrow3⋮\left(n+1\right)\)

\(\Rightarrow n+1\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\)

\(\Rightarrow n\in\left\{-4,-2,0,2\right\}\)

b,Để A có GTNN thì \(\frac{3}{n+1}\) lớn nhất

\(\Rightarrow n+1\) bé nhất và n+1>0

\(\Rightarrow n+1=1\Rightarrow n=0\)

Nên GTNN của A=-1