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

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

Đề bài này có sai không vậy ?

sai đề bài oidf               

k sai đâu bn

`Q=(a^3-3a^2+3a-1)/(a^2-1)`
`a)ĐK:a^2-1 ne 0<=>a ne +-1`
`Q=(a^3-3a^2+3a-1)/(a^2-1)`
`=(a-1)^3/((a-1)(a+1))`
`=(a-1)^2/(a+1)`
`b)|a|=5`
`<=>`  \(\left[ \begin{array}{l}a=5\\a=-5\end{array} \right.\) 
`<=>`  \(\left[ \begin{array}{l}Q=\dfrac{(5-1)^2}{5+1}=\dfrac83\\Q=\dfrac{(-5-1)^2}{-5+1}=-9\end{array} \right.\) 

`a)D` xác định `<=>a-1 ne 0<=>a ne 1`

`b)` Với `a ne 1` có:

`D=([a-1]/[a^2+a+1]-[1-3a+a^2]/[(a-1)(a^2+a+1)]-1/[a-1]).[1-a]/[a^2+1]`

`D=[(a-1)^2-1+3a-a^2-a^2-a-1]/[(a-1)(a^2+a+1)].[-(a-1)]/[a^2+1]`

`D=[a^2-2a+1-1+3a-a^2-a^2-a-1]/[(-a^2-1)(a^2+a+1)]`

`D=[-a^2-1]/[(-a^2-1)(a^2+a+1)]=1/[a^2+a+1]`

`c)` Với `a ne 1` có:

`1/D=1/[1/[a^2+a+1]]=a^2+a+1=(a+1/2)^2+3/4`

Vì `(a+1/2)^2 >= 0 AA a ne 1`

   `=>(a+1/2)^2+3/4 >= 3/4 AA a ne 1`

  Hay `1/D >= 3/4 AA a ne 1=>1/D  _[mi n]=3/4`

Dấu "`=`" xảy ra `<=>a=-1/2` (t/m).

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) Rút gọn

\(Q=\dfrac{a^3-3a^2+3a-1}{a^2-1}\)

= \(\dfrac{a^3-1-3a^2+3a}{\left(a-1\right)\left(a+1\right)}\)

= \(\dfrac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}\)

= \(\dfrac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}\)

= \(\dfrac{\left(a-1\right)^2}{a+1}\)

b)

Tìm giá trị của Q khi |a|=5

**Với a = 5 ta có:

Q= \(\dfrac{\left(5-1\right)^2}{5+1}=\dfrac{4^2}{6}=\dfrac{16}{6}=\dfrac{8}{3}\)

** Với a= -5 ta có:

Q= \(\dfrac{\left(-5-1\right)^2}{-5+1}=\dfrac{\left(-6\right)^2}{-4}=\dfrac{36}{-4}=-9\)

\(\dfrac{a^3-3a^2+3a-1}{a^2-1}=\dfrac{\left(a^3-1\right)-\left(3a^2-3a\right)}{\left(a+1\right)\left(a-1\right)}\)\(\dfrac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}=\dfrac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}=\dfrac{\left(a-1\right)\left(a-1\right)^2}{\left(a-1\right)\left(a+1\right)}\)\(\dfrac{\left(a-1\right)^2}{a+1}\)

Khi |a|=5

a=5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(5-1\right)^2}{5-1}=4\)

a=-5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(-5-1\right)^2}{-5-1}=-6\)

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 Max_A = 1 khi x = 2

•๖ۣۜIηεqυαℓĭтĭεʂ•ッᶦᵈᵒᶫ★T&T★ Idol cho em hỏi là, cái chỗ \(\left(a-2\right)^2\ge0\) thì tại sao Khi đó: \(\frac{4a}{a^2+4}\le1\)

Mong Idol pro giải thích hộ em chỗ này :((

À dạ thôi oke, em hiểu rồi((: