cho biểu thức P=\(\left(\dfrac{a}{3a^2-6a}+\dfrac{2a-3}{2a^2-a^3}\right).\dfrac{6a}{a^2-6a+9}\)
a.rút gọn P
b.tìm giá trị của A để P>0
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NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
A
4 tháng 1 2018
a)
Để B được xác định khi:
*\(2a^2+6a\ne0\Rightarrow2a\left(a+3\right)\ne0\)
=>\(a\ne0\) và \(a+3\ne0\Rightarrow a\ne-3\)
*a2-9\(\ne\)0
=>(a+9)(a-9)\(\ne\)0
=> a+9\(\ne\)0 và a-9\(\ne\)0
=> a \(\ne\)-9 và a\(\ne\)9
Vậy để B được xác định khi a\(\in\){-9;-3;0;9}
b)
\(\dfrac{\left(a+3\right)^2}{2a^2+6a}\cdot\left(1-\dfrac{6a-18}{a^2-9}\right)\)
=\(\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\left(1-\dfrac{6\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)\)
=\(\dfrac{a+3}{2a}\cdot\left(1-\dfrac{6}{a+3}\right)\)
=\(\dfrac{a+3}{2a}\cdot\left(\dfrac{a+3-6}{a+3}\right)\)
=\(\dfrac{a+3}{2a}\dfrac{a-3}{a+3}\)
=\(\dfrac{a-3}{2a}\)
c) Ta có B=0
=>\(\dfrac{a-3}{2a}=0\\ \Rightarrow a-3=0\\ \Rightarrow a=3\)
Vậy B=0 khi a=3
d)Ta có B=1
\(\Rightarrow\dfrac{a-3}{2a}=1\\ \Rightarrow a-3=2a\\ \Rightarrow a-2a=3\\ \Rightarrow-a=3\\ \Rightarrow a=-3\left(KTMDK\right)\)
23 tháng 12 2018
a)
Để B được xác định khi:
*2a2+6a≠0⇒2a(a+3)≠0
=>a≠0 và a+3≠0⇒a≠−3
*a2-9≠0
=>(a+9)(a-9)≠0
=> a+9≠0 và a-9≠0
=> a ≠-9 và a≠9
Vậy để B được xác định khi a∈{-9;-3;0;9}
b)
(a+3)22a2+6a⋅(1−6a−18a2−9)
=(a+3)22a(a+3).(1−6(a−3)(a−3)(a+3))
=a+32a⋅(1−6a+3)
=a+32a⋅(a+3−6a+3)
=a+32aa−3a+3
=a−32a
c) Ta có B=0
=>a−32a=0⇒a−3=0⇒a=3
Vậy B=0 khi a=3
d)Ta có B=1
Y
10 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\left(dk:a\ne1,a\ne-1\right)\)
\(=-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right):\dfrac{2a+1}{\left(a-1\right)\left(a+1\right)}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}.\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)
\(=-\dfrac{2}{2a+1}\)
H
11 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a-1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{3a+1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)\(=-\left(\dfrac{a^2-2a+1-\left(a^2+a\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{2}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =\dfrac{-2.\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a+1\right).\left(2a+1\right)}\\ =-\dfrac{2}{2a+1}\)
__
\(-\dfrac{2}{2a+1}=\dfrac{3}{a-1}\\ \Leftrightarrow-2\left(a-1\right)=3\left(2a+1\right)\\ \Leftrightarrow-2a+2-6a-3=0\\ \Leftrightarrow-8a-1=0\\ \Leftrightarrow-8a=1\\ \Leftrightarrow a=-\dfrac{1}{8}\)
6 tháng 2 2022
a) \(P=\dfrac{\sqrt{a}\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(P=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(P=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(P=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(P=a-\sqrt{a}\)
b) Với a > 1 thì \(a>\sqrt{a}\) , do đó \(P=a-\sqrt{a}>0\), suy ra \(\left|P\right|=P\)
c) \(A=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy A nhỏ nhất bằng \(-\dfrac{1}{4}\) khi cà chỉ khi \(\sqrt{a}=\dfrac{1}{2}\) hay \(a=\dfrac{1}{4}\)
6 tháng 2 2022
a: \(P=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: a>1 nên P>0
\(\Leftrightarrow P=\left|P\right|\)
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 \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)