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\(a^3+6=-3a-2a^2\)
\(\Leftrightarrow a^3+2a^2+6+3a=0\)
\(\Leftrightarrow a^2\left(a+2\right)+3\left(a+2\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(a^2+3\right)=0\)
\(\Leftrightarrow a+2=0\left(do.a^2+3>0\right)\)
<=>a=-2
thay a=-2 vào biểu thức ta được \(A=\frac{-2-1}{-2+3}=\frac{-3}{1}=-3\)
Ta có : a3+6=-3a-2a2
<=> a3+6+3a+2a2=0
<=>(a3+2a2)+(3a+6)=0
<=>a2(a+2)+3(a+2)=0
<=>(a2+3)(a+2)=0
\(\hept{\begin{cases}a^2+3=0\\a+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a^2=-3\\a=-2\end{cases}\Leftrightarrow}\hept{\begin{cases}a\in\varnothing\\a=-2\end{cases}}}\)
Thay a=-2 vào biểu thức :
=> A= \(\frac{-2-2}{-2+3}=\frac{-4}{1}=-4\)
a3+6= -3a-2a2.
->a=-2
\(\Leftrightarrow A=\frac{-2-1}{-2+3}=\frac{-3}{1}=-3\)
vậy A=-3
ĐKXD: a+3 khác 0 => a khác -3
Ta có a^3+6+3a+2a^2=0
<=> a^2(a+2) + 3(a+2)=0
<=> (a+2)(a^2+3)=0
=> a+2=0 <=> a= -2
Suy ra
a-1/a+3= -2-1/-2+3=-3/1=-3
a: ĐKXĐ: a∉{-1/3;-3}
\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)
=>\(2\left(3a+1\right)\left(a+3\right)=\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)\)
=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)
=>\(6a^2+20a+6=6a^2-6\)
=>20a=-12
=>a=-3/5(nhận)
b: ĐKXĐ: a∉{5/2;2/3}
\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)
=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)
=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)
=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)
=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)
=>6a-4=2a-5
=>4a=-1
=>a=-1/4(nhận)
c: ĐKXĐ: a<>-3
\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)
=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)
=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)
=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)
=>23a+1=16(a+3)=16a+48
=>7a=47
=>a=47/7(nhận)
\(a^3+6=-3a-2a^2\)
\(\Leftrightarrow a^3+6+3a+2a^2=0\)
\(\Leftrightarrow a^2.\left(a+2\right)+3.\left(a+2\right)=0\)
\(\Leftrightarrow\left(a^2+3\right).\left(a+2\right)=0\Leftrightarrow a+2=0\Leftrightarrow a=-2\left(\text{vì }a^2+3\ge3\right)\)
Thay a=-2, vào A ta có:
\(A=\frac{-2-1}{-2+3}=-3\)
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.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
\(\left|a^2-3a+1\right|=1\)
=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)
=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)
ĐKXĐ: a<>2
=>a∈{0;1;3}
\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)
\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)
\(=2a^2-8a+1\)
Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)
Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)
Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)