Question

[Nguyễn Bích Hạnh]

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

tim a thuoc z de m thuoc z

tim a de m=7.tim a de m>0

Answer 1

[Trần Thanh Phương]

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 : \(a-1>0\Leftrightarrow a>1\)

Answer 2

[Trần Thanh Phương]

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 \(\Leftrightarrow a+\frac{1}{2}=0\Leftrightarrow a=\frac{-1}{2}\)