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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, \(M=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1}{3a}-a-1\right)\right]:\frac{a-1}{a}\)ĐK : \(a\ne\pm1;0\)
\(=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1-3a^2-3a}{3a}\right)\right]:\frac{a-1}{a}\)
\(=\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{-3a^2-2a+1}{3a}\right)\right]:\left(\frac{a-1}{a}\right)\)
\(=\left[\frac{2}{3a}+\frac{2}{a+1}.\frac{\left(a+1\right)\left(3a-1\right)}{3a}\right]:\left(\frac{a-1}{a}\right)\)
\(=\left(\frac{2}{3a}+\frac{2\left(3a-1\right)}{3a}\right):\left(\frac{a-1}{a}\right)=\frac{2a}{a-1}\)
b, Để P nguyên \(\frac{2a}{a-1}=\frac{2\left(a-1\right)+2}{a-1}=2+\frac{2}{a-1}\)
Vì 2 nguyên nên \(\frac{2}{a-1}\)cũng phải nguyên
\(\Rightarrow a-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| a - 1 | 1 | -1 | 2 | -2 |
| a | 2 ( tm ) | 0 ( tm ) | 3 (tm ) | -1 (tm) |
c, Ta có : \(P\le1\Rightarrow\frac{2a}{a-1}\le1\Leftrightarrow\frac{2a}{a-1}-1\le0\)
\(\Leftrightarrow\frac{a+1}{a-1}\le0\)mà a + 1 > a - 1
\(\hept{\begin{cases}a+1\ge0\\a-1\le0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ge-1\\a\le1\end{cases}\Leftrightarrow-1\le}a\le1}\)
Kết hợp với đk vậy \(-1< a< 1;a\ne0\)thì \(P\le1\)
\(P=\frac{1}{a^2+a+1}\) ( với a khác 1 )
=> \(\frac{1}{P}=a^2+a+1=a^2+2.a.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(=\left(a+\frac{1}{2}\right)^2+\frac{3.}{4}\ge\frac{3}{4}\) vì \(\left(a+\frac{1}{2}\right)^2\ge0\forall a\)
Dấu "=" xảy ra <=> \(\left(a+\frac{1}{2}\right)^2=0\Leftrightarrow a=-\frac{1}{2}\)( thỏa mãn )
Vậy GTNN của \(\frac{1}{P}=\frac{3}{4}\)đạt tại a = - 1/2.
a) ĐKXĐ : \(\hept{\begin{cases}a\ne0\\a\ne-1\\a\ne1\end{cases}}\)
Khi đó P = \(\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1}{3a}-a-1\right)\right]:\frac{a-1}{a}\)
\(=\left[\frac{2}{3a}-\frac{2}{a+1}.\frac{a+1}{3a}+\frac{2}{a+1}.\left(a+1\right)\right]:\frac{a-1}{a}\)
\(=\left(\frac{2}{3a}-\frac{2}{3a}+2\right):\frac{a-1}{a}=2:\frac{a-1}{a}=\frac{2a}{a-1}\)
b) Ta có P = \(\frac{2a}{a-1}=\frac{2a-2+2}{a-1}=2+\frac{2}{a-1}\)
\(P\inℤ\Leftrightarrow2⋮a-1\Leftrightarrow a-1\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
<=> \(a\in\left\{2;3;0;-1\right\}\)
c) Để P \(\le1\)
<=> \(\frac{2a}{a-1}\le1\)
<=> \(\frac{a+1}{a-1}\le0\)
Xét 2 trường hợp
TH1 : \(\hept{\begin{cases}a+1\ge0\\a-1\le0\end{cases}}\Leftrightarrow-1\le a\le1\)
Kết hợp điều kiện => -1 < a < 1 (a \(\ne0\))
TH2 : \(\hept{\begin{cases}a+1\le0\\a-1\ge0\end{cases}}\Leftrightarrow a\in\varnothing\)
Vậy - 1 < a < 1 (a \(\ne0\))