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a) \(ĐKXĐ:\hept{\begin{cases}a\ne\pm2\\a\ne1\\a\ne0\end{cases}}\)
\(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)
\(\Leftrightarrow A=\frac{8a-4a^2+8a^2}{\left(2-a\right)\left(2+a\right)}:\frac{a-3-2a+4}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a}{2-a}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2\left(a-2\right)}{\left(a-2\right)\left(a-1\right)}\)
\(\Leftrightarrow A=\frac{4a^2}{a-1}\)
b) Để A nhận giá trị nguyên
\(\Leftrightarrow\frac{4a^2}{a-1}\inℤ\)
\(\Leftrightarrow4a^2⋮a-1\)
\(\Leftrightarrow4\left(a^2-1\right)+4⋮a-1\)
\(\Leftrightarrow4\left(a-1\right)\left(a+1\right)+4⋮a-1\)
\(\Leftrightarrow4⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)
Ta sẽ loại các giá trị ở đkxđ
Vậy để \(A\inℤ\Leftrightarrow a\in\left\{2;-1;3;-3;5\right\}\)
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, ĐKXĐ : \(\left\{{}\begin{matrix}a\pm2\ne0\\a\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a\ne\pm2\\a\ne0\end{matrix}\right.\)
Ta có : \(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)
=> \(A=\left(\frac{4a}{2+a}+\frac{8a^2}{\left(2-a\right)\left(2+x\right)}\right):\left(\frac{a-3}{a\left(a-2\right)}-\frac{2}{a}\right)\)
=> \(A=\left(\frac{4a\left(2-a\right)}{\left(2+a\right)\left(2-a\right)}+\frac{8a^2}{\left(2-a\right)\left(2+x\right)}\right):\left(\frac{a-3}{a\left(a-2\right)}-\frac{2\left(a-2\right)}{a\left(a-2\right)}\right)\)
=> \(A=\left(\frac{8a-4a^2+8a^2}{\left(2+a\right)\left(2-a\right)}\right):\left(\frac{a-3-2a+4}{a\left(a-2\right)}\right)\)
=> \(A=\left(\frac{4a}{2-a}\right):\left(\frac{1-a}{a\left(a-2\right)}\right)\)
=> \(A=\frac{4a^2\left(a-2\right)}{\left(2-a\right)\left(1-a\right)}=\frac{-4a^2}{1-a}=\frac{4a^2}{a-1}\)