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a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
`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,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) ĐKXĐ: a2-1 ≠0 ⇔ (a-1)(a+1)≠0 ⇔\(\left[{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)
b) A=\(\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\) , a≠1, -1
=\(\dfrac{2a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2-a\left(a-1\right)+a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2-a^2+a+a^2+a}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2+2a}{\left(a-1\right)\left(a+1\right)}\) =\(\dfrac{2a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\) =\(\dfrac{2a}{a-1}\)
vậy A =\(\dfrac{2a}{a-1}\) với a≠1,-1.
c) Có:A= \(\dfrac{2a}{a-1}\) = \(\dfrac{2a-2+2}{a-1}=\dfrac{2\left(a-1\right)+2}{a-1}=2+\dfrac{2}{a-1}\)
Để a∈Z thì a-1 ∈ Z ⇒ (a-1) ∈ Ư(2) =\(\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
Vậy để biểu thức A có giá trị nguyên thì a∈\(\left\{2;0;3\right\}\)
a) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)
b) Ta có: \(A=\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\)
\(=\dfrac{2a^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}+\dfrac{a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a^2-a^2+a+a^2+a}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a^2+2a}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a}{a-1}\)
c) Để A nguyên thì \(2a⋮a-1\)
\(\Leftrightarrow2a-2+2⋮a-1\)
mà \(2a-2⋮a-1\)
nên \(2⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(2\right)\)
\(\Leftrightarrow a-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow a\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(a\in\left\{0;2;3\right\}\)
Vậy: Để A nguyên thì \(a\in\left\{0;2;3\right\}\)