\(3^{x^2-x-6}

a)   \(x\in\left(\frac{1}{2}-\frac{\sqrt{25\ln3+8\ln2}}{2\sqrt{\ln3}};\frac{\sqrt{25\ln3+8\ln2}}{2\sqrt{\ln3}}+\frac{1}{2}\right)\)

b)   3^{x2 - x - 6} - 1 = 0

x = -2

x = 3

a: =>\(\dfrac{x^2+2x-13-x+1}{x-1}< 0\)

=>\(\dfrac{x^2+x-12}{x-1}< 0\)

=>\(\dfrac{\left(x+4\right)\left(x-3\right)}{x-1}< 0\)

=>1<x<3 hoặc x<-4

b: =>\(\dfrac{3x^2+4x-3x-4}{x-1}< 3\)

=>3x+4<3

=>3x<-1

=>x<-1/3

c: TH1: 2x^2-3x+1>0 và x+2>0

=>(2x-1)(x-1)>0 và x+2>0

=>x>1

TH2: (2x-1)(x-1)<0 và x+2<0

=>x<-2 và 1/2<x<1

=>Loại

1) ĐKXĐ: \(\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

ta có: (-6).\(\sqrt{6x^2-18x+12}\) > \(6x^2-18x-60\)

⇔ \(6x^2-18x+12\) + \(2.3.\sqrt{6x^2-18x+12}+9-81\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+3\right)^2-9^2\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+12\right).\left(\sqrt{6x^2-18x+12}-6\right)\) > 0

⇔ \(\sqrt{6x^2-18x+12}-6\) > 0

⇔ \(\sqrt{6x^2-18x+12}>6\)

⇔\(6x^2-18x+12>36\)

⇔ \(6x^2-18x-24>0\)

⇔\(\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\)

đối chiếu ĐKXĐ ban đầu ta được: x ϵ (-∞;-1) \(\cup\)(4;+∞)

b) ĐKXĐ: \(\forall x\) ϵ R

\(\left(x-2\right)\sqrt{x^2+4}-\left(x-2\right)\left(x+2\right)\le0\)

⇔\(\left(x-2\right)\left(\sqrt{x^2+4}-x-2\right)\le0\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\\sqrt{x^2+4}-x-2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\\sqrt{x^2+4}-x-2\ge0\end{matrix}\right.\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x^2+4\le x^2+4x+4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x^2+4\ge x^2+4x+4\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)

Đối chiếu ĐKXĐ ta được x ϵ ( -∞;0) \(\cup\)( 2; +∞)

ĐKXĐ: \(x^2+x-1\ge0\)

\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2+b^2>3ab\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)

\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)

\(bpt\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\4-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\4-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x>4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1\le x< 4\)

Vậy .......

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\4-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\4-x< 0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x>4\end{matrix}\right.\end{matrix}\right.\)

Vậy....

$\begin{cases}|x^2-5x+4|>x-1\\x>1\\\end{cases}$

$\to \begin{cases}(x^2-5x+4)^2>(x-1)^2\\x>1\\\end{cases}$

$\to \begin{cases}(x-1)^2(x-4)^2>(x-1)^2\\x>1\\\end{cases}$

$\to \begin{cases}(x-1)^2[(x-4)^2-1]>0\\x>1\\\end{cases}$

$\to \begin{cases}(x-4)^2-1>0\\x>1\\\end{cases}$

$\to \begin{cases}(x-5)(x-3)>0\\x>1\\\end{cases}$

$\to \begin{cases}\left[ \begin{array}{l}x>5\\x<3\end{array} \right.\\x>1\\\end{cases}$

$\to \left[ \begin{array}{l}1<x<3\\x>5\end{array} \right.$

Vậy bất phương trình có tập nghiệm $S=(1,3]∩(5,∞]$

Giúp vs ạ mk đag cần

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4x2 hay là 4x² vậy bạn ?