a: \(Q=\left(\dfrac{a^2+4a+4-a^2+4a-4+4a^2}{\left(a-2\right)\left(a+2\right)}\right):\dfrac{a\left(a-3\right)}{5a\left(2-a\right)}\)

\(=\dfrac{4a^2+8a}{\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-5\left(a-2\right)}{a-3}\)

\(=\dfrac{-20a}{a-3}\)

b: Q chia hết cho 20 thì a/a-3 là số nguyên

=>\(a-3\in\left\{1;-1;3;-3\right\}\)

=>a=4 hoặc a=6

\(a,Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)^2}{a+1}.\)

b, ta có : \(/a/=5\Rightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

thay a = -5 vào Q 

\(\Rightarrow Q=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-4}=-9\)

thay a = 5 vào Q 

\(\Rightarrow Q=\frac{\left(5-1\right)^2}{5+1}=\frac{16}{6}=\frac{8}{3}\)

KL : Q = 8/3 tại x=5

\(\text{Đ}K\text{X}\text{Đ}:a\ne1\)

a) Ta có: \(Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(a+1\right)}\)

Vậy ....

b) Ta có: \(\left|a\right|=5\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

Với a=5 ta có: \(Q=\frac{\left(5-1\right)^2}{5+1}=\frac{16}{6}=\frac{8}{3}\)

Với a=-5 ta có: \(Q=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-4}=-9\)

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 

\(\Rightarrow a-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

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 \(P\le1\)

cảm ơn bạn

Đề bài này có sai không vậy ?

sai đề bài oidf               

k sai đâu bn

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 \(\ne0\))

\(P=\frac{1}{a^2+a+1}\) ( 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:\(x\ge\frac{1}{3};x\ne1\)

b)\(P=\frac{3a+\sqrt{9a-3}-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+6+\sqrt{9a-3}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)