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a: ĐKXĐ: a<>3; a<>-3; a<>-1
b: \(P=\dfrac{2a^2-3a+3a+9-2a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)
\(=\dfrac{6}{\left(a+3\right)\left(a+1\right)}\)
c: |a|=2
=>a=2 hoặc a=-2
Khi a=-2 thì \(P=\dfrac{6}{\left(-2+3\right)\left(-2+1\right)}=-6\)
Khi a=2 thì \(P=\dfrac{6}{\left(2+3\right)\left(2+1\right)}=\dfrac{6}{5\cdot3}=\dfrac{2}{5}\)
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:
| a-1 | 1 | -1 | 2 | -2 |
| a | 2 | 0 | 3 | -1 |
| Thử lại | TM | TM | TM | ko TM(vì a≠-1 |
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 \(a\in\left\{0;2;3\right\}\)
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\left(dk:a\ne1,a\ne-1\right)\)
\(=-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right):\dfrac{2a+1}{\left(a-1\right)\left(a+1\right)}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}.\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)
\(=-\dfrac{2}{2a+1}\)
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a-1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{3a+1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)\(=-\left(\dfrac{a^2-2a+1-\left(a^2+a\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{2}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =\dfrac{-2.\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a+1\right).\left(2a+1\right)}\\ =-\dfrac{2}{2a+1}\)
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\(-\dfrac{2}{2a+1}=\dfrac{3}{a-1}\\ \Leftrightarrow-2\left(a-1\right)=3\left(2a+1\right)\\ \Leftrightarrow-2a+2-6a-3=0\\ \Leftrightarrow-8a-1=0\\ \Leftrightarrow-8a=1\\ \Leftrightarrow a=-\dfrac{1}{8}\)
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
Lời giải:
a) ĐKXĐ: $a\neq 0; a\neq 3; a\neq 2$
\(P=\left[\frac{a}{3a(a-2)}-\frac{2a-3}{a^2(a-2)}\right].\frac{6a}{(a-3)^2}=\left[\frac{a^2}{3a^2(a-2)}-\frac{6a-9}{3a^2(a-2)}\right].\frac{6a}{(a-3)^2}=\frac{a^2-6a+9}{3a^2(a-2)}.\frac{6a}{(a-3)^2}=\frac{(a-3)^2}{3a^2(a-2)}.\frac{6a}{(a-3)^2}=\frac{2}{a(a-2)}\)
b)
Để $P>0\Leftrightarrow \frac{2}{a(a-2)}>0\Leftrightarrow a(a-2)>0$
$\Leftrightarrow a>2$ hoặc $a< 0$
Kết hợp với ĐKXĐ suy ra $(a>2; a\neq 3)$ hoặc $a< 0$
ĐKXĐ: \(a\notin\left\{0;2\right\}\)
a) Ta có: \(P=\left(\dfrac{a}{3a^2-6a}+\dfrac{2a-3}{2a^2-a^3}\right)\cdot\dfrac{6a}{a^2-6a+9}\)
\(=\left(\dfrac{a}{3a\left(a-2\right)}+\dfrac{2a-3}{a^2\left(2-a\right)}\right)\cdot\dfrac{6a}{a^2-6a+9}\)
\(=\left(\dfrac{a^2}{3a^2\cdot\left(a-2\right)}-\dfrac{3\left(2a-3\right)}{3a^2\cdot\left(a-2\right)}\right)\cdot\dfrac{6a}{\left(a-3\right)^2}\)
\(=\dfrac{a^2-6a+9}{3a^2\cdot\left(a-2\right)}\cdot\dfrac{6a}{\left(a-3\right)^2}\)
\(=\dfrac{\left(a-3\right)^2}{3a^2\left(a-2\right)}\cdot\dfrac{6a}{\left(a-3\right)^2}\)
\(=\dfrac{2}{a\left(a-2\right)}\)
b) Để P>0 thì \(\dfrac{2}{a\left(a-2\right)}>0\)
mà 2>0
nên \(a\left(a-2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\a-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\a-2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\a>2\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\a< 2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
Vậy: Để P>0 thì \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)
\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)
\(=\dfrac{-4a^2-2a+1}{4a^2}\)
a) \(ĐKXĐ:a\ne\pm1\)
b) \(P=\left(\dfrac{a+1}{2a-2}+\dfrac{1}{2-2a^2}\right)\cdot\dfrac{2a+2}{a+2}\)
\(=\left(\dfrac{a+1}{2\left(a-1\right)}+\dfrac{1}{2\left(1-a^2\right)}\right)\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\left(\dfrac{a+1}{2\left(a-1\right)}-\dfrac{1}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)-1}{2\left(a-1\right)\left(a+1\right)}\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\dfrac{a^2-1-1}{\left(a-1\right)\left(a+2\right)}\)
\(=\dfrac{a^2-2}{a^2+a-2}\)
Khi a = 2 thì :
\(P=\dfrac{2^2-2}{2^2+2-2}=\dfrac{2}{4}=\dfrac{1}{2}\)
p/s: check lại hộ tui nhá =)))
thêm cho mình đkxđ : a \(\ne\) - 2