\(\dfrac{a-2}{8a^2-8a}\)+\(\dfrac{a+1}{6a^3-6a^2}\)
K
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NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 2 2019
\(a=\dfrac{13}{\sqrt{\left(4+\sqrt{3}\right)^2}}=\dfrac{13}{4+\sqrt{3}}=4-\sqrt{3}\Rightarrow\sqrt{3}=4-a\)
\(\Rightarrow3=16-8a+a^2\Rightarrow a^2-8a+13=0\)
\(A=\dfrac{a^2\left(a^2-8a+13\right)+2a^3-15a^2+18a+23}{a^2-8a+13+2}\)
\(A=\dfrac{2a\left(a^2-8a+13\right)+a^2-8a+13+10}{2}\)
\(A=\dfrac{10}{2}=5\)
N
28 tháng 6 2018
a) \(N=8a^3-27b^3\)
\(=\left(2a\right)^3-\left(3b\right)^3\)
\(=\left(2a-3b\right)^3+18ab\left(2a-3b\right)\)
\(=5^3+18\cdot12\cdot5\)
\(=125+1080=1205\)
b) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+2ab+b^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a+b\right)^2\)
\(=\left(a+b\right)^3+3ab\left(a+b\right)\left(a+b-1\right)\)
\(=1^3+3ab\cdot1\cdot0\)
\(=1\)
28 tháng 6 2018
a ) \(N=8a^3-27b^3\)
\(\Leftrightarrow N=\left(2a-3b\right)\left(4x^2+6ab+9b^2\right)\)
\(\Leftrightarrow N=5\left(4x^2+9b^2+72\right)\)
Ta có : \(2a-3b=5\)
\(\Leftrightarrow4a^2+9b^2=25+6ab\)
Thay vào ta được : \(N=5\left(25+6ab+72\right)=845\)
b ) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(\Leftrightarrow K=\left(a+b\right)^3-3ab\left(a+b\right)+6a^2b^2\left(a+b\right)+3ab\left(a+b\right)^2-6a^2b^2\)
\(\Leftrightarrow K=1-3ab+6a^2b^2+3ab-6a^2b^2=1\)
c ) \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)
\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)
\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)
\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)
\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)
AH
Akai Haruma
Giáo viên
15 tháng 2 2021
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 $a< 0$
15 tháng 2 2021
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 \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
HT
3
19 tháng 8 2016
thay a = 2 => 2 + 1 = 3
vậy a = 3
còn lại tương tự!! 465464565775685687435425345643645654657657
HB
21 tháng 10 2020
\(8a^3-6a^2-1+3a\)
\(=\left(2a-1\right)^3\)
#Satan_Dilys
#5:20_21/10/2020
19 tháng 8 2018
a) điều kiện xác định : \(a\ge0;a\ne1\)
ta có : \(P=\dfrac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\dfrac{\sqrt{a}-2}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+2}-1\)
\(\Leftrightarrow P=\dfrac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}-2}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}+2}\) \(\Leftrightarrow P=\dfrac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\) \(\Leftrightarrow P=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)để \(\left|P\right|=1\Leftrightarrow\left|\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right|=1\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=1\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-1=0\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{\sqrt{a}-1}=0\\\dfrac{2\sqrt{a}}{\sqrt{a}-1}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2=0\left(vôlí\right)\\2\sqrt{a}=0\end{matrix}\right.\Rightarrow a=0\)
vậy \(a=0\)
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