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<=> (12²x²+2.12.7x + 7²).(6x²+7x+2) = 3
<=> [24.(6x² +7x +2) +1].(6x² +7x +2) =3
đặt: a= 6x² +7x +2
<=> (24a+1).a = 3
=> a=...
![](https://rs.olm.vn/images/avt/0.png?1311)
x3-6x2+11x-6=0
⇔x3-x2-5x2+5x+6x-6=0
⇔(x3-x2)-(5x2-5x)+(6x-6)=0
⇔x2(x-1)-5x(x-1)+6(x-1)=0
⇔(x-1)(x2-5x+6)=0
⇔(x-1)(x2-2x-3x+6)=0
⇔(x-1)[(x2-2x)-(3x-6)]=0
⇔(x-1)[x(x-2)-3(x-2)]=0
⇔(x-1)(x-2)(x-3)=0
=>\(\left\{{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy S={1;2;3}
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Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)
\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)
\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)
Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(6\left(a^2-2\right)+7a-36=0\)
\(\Leftrightarrow6a^2+7a-48=0\)
Nghiệm xấu
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\(x^2+\left(16-x\sqrt{3}\right)^2=4\left(12-x\right)^2\)
\(\Leftrightarrow x^2+256-32\sqrt{3}x+3x^2=4\left(144-24x+x^2\right)\)
\(\Leftrightarrow4x^2-32\sqrt{3}x+256=576-96x+4x^2\)
\(\Leftrightarrow4x^2-4x^2-32\sqrt{3}x+96x+256-576=0\)
\(\Leftrightarrow\left(96-32\sqrt{3}\right)x-320=0\)
\(\Leftrightarrow\left(96-32\sqrt{3}\right)x=320\)
\(\Leftrightarrow x=\frac{320}{96-32\sqrt{3}}=\frac{15+5\sqrt{3}}{3}\)
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\(a,x^2< 1=1^2=>x< 1\) thỏa mãn bất phương trình
\(b,2x+5\ge7=>2x\ge7-5=2=>x\ge1\) thỏa mãn bất phương trình
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x+2\left(5-x\right)=-11\)
\(\Leftrightarrow3x+10-2x=-11\)
\(\Leftrightarrow x=-21\)
b) \(3x^2-3x\left(x-2\right)=36\)
\(\Leftrightarrow3x^2-3x^2+6x=36\)
\(\Rightarrow x=6\)
\(\left(x^2+6x+5\right)\left(x+4\right)\left(x+2\right)=40\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt \(x^2+6x+5=t\) ,ta có:
\(t\left(t+3\right)=40\)
\(\Leftrightarrow t^2+3t-40=0\)
\(\Leftrightarrow t^2+8t-5t-40=0\)
\(\Leftrightarrow\left(t+8\right)\left(t-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-8\\t=5\end{matrix}\right.\)
Với t = -8
\(x^2+6x+5=-8\)
\(\Leftrightarrow x^2+6x+13=0\) ( vô lý vì \(x^2+6x+13>0\forall x\) )
Với t = 5
\(x^2+6x+5=5\)
\(\Leftrightarrow x^2+6x=0\)
\(\Leftrightarrow x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy ............................