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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 )
\(x+2\sqrt{2x^2}+2x^3=0\)
\(\Leftrightarrow x+2x\sqrt{2}+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}+2x^2\right)=0\)
\(\Leftrightarrow x=0\) ( Vì \(1+2\sqrt{2}+2x^2>0\) )
Tìm x biết :
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(1+\sqrt{2}x\right)^2=0\)
TH1 : x=0
TH2 : \(\left(1+\sqrt{2}x\right)^2=0\)
\(1+\sqrt{2}x=0\)
\(x=\frac{-1}{\sqrt{2}}\)
2x3 + 3x2 + 6x + 5 = 02
<=> 2x3 + x2 + 5x + 2x2 + x + 5 = 0
<=> x(2x2 + x + 5) + (2x2 + x + 5) = 0
<=> (2x2 + x + 5)(x + 1) = 0
<=> x + 1 = 0 (vì 2x2 + x + 5 \(\ge\) 4,875 > 0 \(\forall\) x)
<=> x = - 1
Vậy tập nghiệm của pt là \(S=\left\{-1\right\}\)
b) 4x4 + 12x3 + 5x2 - 6x - 15 = 0
<=> 4x4 + 10x3 + 2x3 + 5x2 - 6x - 15 = 0
<=> 2x3(2x + 5) + x2(2x + 5) - 3(2x + 5) = 0
<=> (2x + 5)(2x3 + x2 - 3) = 0
<=> (2x + 5)(2x3 - 2x2 + 3x2 - 3) = 0
<=> (2x + 5)(x - 1)(2x2 + 3x + 3) = 0
<=> (2x + 5)(x - 1)[x2 + (x + 3/2)2 + 3/4]= 0
Mà x2 + (x + 3/2)2 + 3/4 > 0\(\forall x\)
\(\Rightarrow\left[\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=1\end{matrix}\right.\)
Vậy ...
\(\dfrac{90}{x+6}-\dfrac{36}{x}=2\) 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}