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a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2
a) \(ĐKXĐ:\hept{\begin{cases}a\ne1\\a\ne0\end{cases}}\)
\(M=\left(\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right)\div\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left(\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right):\frac{a^2+4}{4a}\)
\(\Leftrightarrow M=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(\Leftrightarrow M=\frac{a^3-3a^2+3a-1-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(\Leftrightarrow M=\frac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a^2}{a^2+4}\)
\(\Leftrightarrow M=\frac{4a^2}{a^2+4}\)
b) Ta có : \(\frac{4a^2}{a^2+4}=\frac{4\left(a^2+4\right)-16}{a^2+4}\)
\(=4-\frac{16}{a^2+4}\)
Để M đạt giá trị lớn nhất
\(\Leftrightarrow\frac{16}{a^2+4}\)min
\(\Leftrightarrow a^2+4\)max
\(\Leftrightarrow a\)max
Vậy để M đạt giá trị lớn nhất thì a phải đạ giá trị lớn nhất.
a) \(a\ne0;a\ne1\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(=\left[\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right]\cdot\frac{4a^2}{a\left(a^2+4\right)}\)
\(=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(=\frac{a^3-1}{a^3-1}\cdot\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
Vậy \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
b) \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
M>0 khi 4a>0 => a>0
Kết hợp với ĐKXĐ
Vậy M>0 khi a>0 và a\(\ne\)1
c) \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
\(M=\frac{4a}{a^2+4}=\frac{\left(a^2+4\right)-\left(a^2-4a+4\right)}{a^2+4}=1-\frac{\left(a-2\right)^2}{a^2+4}\)
Vì \(\frac{\left(a-2\right)^2}{a^2+4}\ge0\forall a\)nên \(1-\frac{\left(a-2\right)^2}{a^2+4}\le1\forall a\)
Dấu "=" <=> \(\frac{\left(a-2\right)^2}{a^2+4}=0\)\(\Leftrightarrow a=2\)
Vậy \(Max_M=1\)khi a=2
\(A=\left(\frac{1-a^3}{a-a^2}+1\right)\cdot\left(\frac{1+a^3}{1+a}-a\right):\frac{\left(1-a^2\right)^3}{1+a}\)
\(=\left(\frac{\left(1-a\right)\cdot\left(1+a+a^2\right)}{a\cdot\left(1-a\right)}+1\right)\cdot\left(\frac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)\(:\frac{\left(1-a\right)^3\cdot\left(1+a\right)^3}{1+a}\)
\(=\left(\frac{1+a+a^2+a}{a}\right)\cdot\left(1-a+a^2-a\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{1+2a+a^2}{a}\cdot\left(1-2a+a^2\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{\left(1+a\right)^2}{a}\cdot\left(1-a\right)^2:\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\text{[}\frac{\left(1+a\right)^2}{a}:\left(1+a\right)^2\text{]}\cdot\text{[}\left(1-a\right)^2:\left(1-a\right)^3\text{]}\)
\(=\frac{1}{a}\cdot\frac{1}{1-a}=\frac{1}{a\left(1-a\right)}=\frac{1}{a-a^2}\)
Để \(A>A^2\Rightarrow\frac{1}{a-a^2}>\frac{1}{\left(a-a^2\right)^2}\)
Có ĐKXĐ : \(\left(a-a^2\right)\ne0\)
Mà \(\left(a-a^2\right)< \left(a-a^2\right)^2\)trừ trường hợp \(\left(a-a^2\right)=1\)
Từ tất cả điều trên suy ra : \(A\)thuộc tất cả các giá trị khác 1 để \(A>A^2\)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
Điều kiện : \(a\ne1\)
\(A=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}+\frac{2a}{a^2+1-a^3-a}\right)-1\)
\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{-a^2-1}{\left(1+a^2\right)\left(1-a\right)}+\frac{2a}{\left(1+a^2\right)\left(1-a\right)}\right)-1\)
\(=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(1+a^2\right)}{\left(a-1\right)^2}-1=\frac{a^2+a+1}{a-1}-1=\frac{a^2+2}{a-1}\)
b) A < 2 \(\Rightarrow\frac{a^2+2}{a-1}< 2\Leftrightarrow\frac{\left(a^2-2a+1\right)+2\left(a-1\right)+3}{a-1}< 2\)
\(\Leftrightarrow a-1+2+\frac{3}{a-1}< 2\Leftrightarrow a-1+\frac{3}{a-1}< 0\)
Đặt t = a-1 , xét :
Nếu t > 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3< 0\) không thỏa mãn vì \(t^2+3>3>0\)
Nếu t < 0 thì \(t+\frac{3}{t}< 0\Leftrightarrow t^2+3>0\) thỏa mãn
Vậy a - 1 < 0 => a < 1 thỏa mãn đề bài