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\(\Leftrightarrow A=\dfrac{\left(x-a\right)^2-\left(x+a\right)^2+3a^2+a}{\left(x-a\right)\left(x+a\right)}\)
\(\Leftrightarrow A=\dfrac{-4ax+3a^2+a}{\left(x-a\right)\left(x+a\right)}\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|\ne a\\4ax=a\left(3a+1\right)\left(1\right)\end{matrix}\right.\)
a) với a=-3
\(\left(1\right)\Leftrightarrow4x=3.\left(-3\right)+1\Rightarrow x=-2\)(NHAN)
b)với a=-1
\(\left(1\right)\Leftrightarrow4x=3.\left(-1\right)+1\Rightarrow x=-\dfrac{2}{4}=-\dfrac{1}{2}\)(NHẬN)
c)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\x=\dfrac{3a+1}{4}=0,5\Rightarrow a=\dfrac{1}{3}\left(nhan\right)\end{matrix}\right.\)
x 2 - x+ y2 -y - 2xy - 7
= ( x2 - 2xy + y2 ) - ( x + y ) -7
= ( x + y )2 - ( x + y ) -7
= ( x + y ) [ ( x + y ) -7]
= ( x + y ) ( x + y - 7 )
\(\dfrac{90}{x+6}-\dfrac{36}{x}=2\) ĐKXĐ: \(x\ne0;x\ne-6\)
\(\Rightarrow90x-36\left(x+6\right)=2x\left(x+6\right)\)
\(\Leftrightarrow90x-36x-216=2x^2+12x\)
\(\Leftrightarrow54x-12x-2x^2-216=0\)
\(\Leftrightarrow42x-2x^2-216=0\)
\(\Leftrightarrow-2\left(x^2-21x+108\right)=0\)
\(\Leftrightarrow x^2-21x+108=0\)
\(\Leftrightarrow x^2-12x-9x+108=0\)
\(\Leftrightarrow x\left(x-12\right)-9\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=9\end{matrix}\right.\)(TMĐK)
S \(=\left\{9;12\right\}\)
\(\dfrac{90}{x+6}-\dfrac{36}{x}=2\)
ĐKXĐ: \(x+6\ne0\) và \(x\ne0\)
MC: x(x+6)
\(\dfrac{90x}{x\left(x+6\right)}-\dfrac{36\left(x+6\right)}{x\left(x+6\right)}=\dfrac{2x\left(x+6\right)}{x\left(x+6\right)}\)
\(\Leftrightarrow90x-36\left(x+6\right)=2x\left(x+6\right)\)
\(\Leftrightarrow90x-36x-216=2x^2+12x\)
\(\Leftrightarrow90x-36x-12x-2x^2-216=0\)
\(\Leftrightarrow42x-2x^2-216=0\)
\(\Leftrightarrow-2\left(x^2-21x+108\right)=0\)
\(\Leftrightarrow x^2-21x+108=0\)
\(\Leftrightarrow x^2-12x-9x+108=0\)
\(\Leftrightarrow\left(x^2-12x\right)-\left(9x-108\right)=0\)
\(\Leftrightarrow x\left(x-12\right)-9\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(x-9\right)=0\)
\(\Leftrightarrow x-12=0\) và \(\Leftrightarrow x-9=0\)
\(\Leftrightarrow x=12\) và \(\Leftrightarrow x=9\) (thỏa ĐK)
Vậy S={12;9}
\(\dfrac{90}{x+6}-\dfrac{36}{x}=2\) ; ( ĐKXĐ: x ≠ -6 và x ≠ 0 )
<=> \(\dfrac{90x}{x\left(x+6\right)}-\dfrac{36\left(x+6\right)}{x\left(x+6\right)}=\dfrac{2x\left(x+6\right)}{x\left(x+6\right)}\)
=> \(90x-36\left(x+6\right)=2x\left(x+6\right)\)
<=> \(90x-36x-216=2x^2+12x\)
<=> \(90x-36x-12x-2x^2-216=0\)
<=> \(-2x^2+42x-216=0\)
<=>\(-2x^2+18x+24x-216=0\)
<=> \(-2x\left(x-9\right)+24\left(x-9\right)=0\)
<=> \(\left(x-9\right)\left(24-2x\right)=0\)
<=>\(\left(x-9\right)2\left(12-x\right)=0\)
Vậy x - 9 =0 hoặc 12 - x =0
1) \(x-9=0\)
<=> x = 9 (thỏa mãn)
2) \(12-x=0\)
<=> x= 12 (thỏa mãn)
Vậy phương trình có tập nghiệm S={ 9; 12}