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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) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
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) \(a\ne0;a\ne1\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(=\left[\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right]\cdot\frac{4a^2}{a\left(a^2+4\right)}\)
\(=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(=\frac{a^3-1}{a^3-1}\cdot\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
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
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
a) \(M=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{3a+a^2-2a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right].\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{\left(a-1\right)^3-\left(1-2a^2+4a\right)+\left(a^2+a+1\right)}{\left(a^2+a+1\right)\left(a-1\right)}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{a^3-3a^2+3a-1-1+2a^2-4a+a^2+a+1}{a^3-1}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{a^3-1}{a^3-1}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{4a}{a^2+4}\)
b) Ta có :
\(\left(a-2\right)^2\ge0\)
\(\Leftrightarrow a^2-4a+4\ge0\)
\(\Leftrightarrow a^2+4\ge4a\)
Dấu " = " xảy ra khi và chỉ khi :
\(\left(a-2\right)^2=0\)
\(\Leftrightarrow a=2\)
Vậy \(Max_M=1\Leftrightarrow a=2\)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)