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1. Đợi chút t tìm cách ngắn gọn.
2. ĐK: \(\left\{{}\begin{matrix}2x^2+8x+6\ge0\\x^2-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge1\\x=-1\end{matrix}\right.\) (*)
BPT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2+8x+5+2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\le\left(2x+2\right)^2\left(1\right)\end{matrix}\right.\)
Giải (1) \(\Leftrightarrow x^2-1-2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\ge0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\right)\ge0\)
TH1: \(\sqrt{x^2-1}=0\Leftrightarrow x=\pm1\) (tm)
TH2: \(x^2-1\ne0\)
\(\Leftrightarrow\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\ge0\)
\(\Leftrightarrow\sqrt{x^2-1}\ge2\sqrt{2x^2+8x+6}\)
\(\Leftrightarrow x^2-1\ge8x^2+32x+24\)
\(\Leftrightarrow7x^2+32x+25\le0\)
\(\Leftrightarrow-\frac{25}{7}\le x\le-1\) kết hợp đk (*) và đk để giải bpt
=>\(x=-1\)
Vậy \(x=\pm1\)
3. ĐK: \(x\ge\frac{4}{5}\)
\(BPT\Leftrightarrow\sqrt{5x-4}-\sqrt{3x-2}+\sqrt{4x-3}-\sqrt{2x-1}>0\)
\(\Leftrightarrow\frac{2x-2}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{2x-2}{\sqrt{4x-3}+\sqrt{2x-1}}>0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{1}{\sqrt{4x-3}+\sqrt{2x-1}}\right)>0\)
\(\Leftrightarrow x-1>0\) \(\Leftrightarrow x>1\)
Vậy \(x>1\)
giải bpt
a) \(x^2-3x-\sqrt{x^2-3x+5}>1\)
b) \(\sqrt[4]{x-\sqrt{x^2-1}}+4\sqrt{x+\sqrt{x^2-1}}-3< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Đặt \(\sqrt{x^2-3x+5}=t>0\)
\(\Leftrightarrow t^2-5-t>1\Leftrightarrow t^2-t-6>0\)
\(\Rightarrow\left[{}\begin{matrix}t>3\\t< -2\left(l\right)\end{matrix}\right.\) \(\Rightarrow\sqrt{x^2-3x+5}>3\)
\(\Leftrightarrow x^2-3x+5>9\Leftrightarrow x^2-3x-4>0\Rightarrow\left[{}\begin{matrix}x>4\\x< -1\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ge1\)
Đặt \(\sqrt[4]{x-\sqrt{x^2-1}}=t>0\Rightarrow\sqrt[4]{x+\sqrt{x^2-1}}=\frac{1}{t}\)
\(\Leftrightarrow t+\frac{4}{t^2}-3< 0\)
\(\Leftrightarrow t^3-3t^2+4< 0\)
\(\Leftrightarrow\left(t+1\right)\left(t-2\right)^2< 0\)
Do \(t>0\Rightarrow t+1>0\Rightarrow VT\ge0\Rightarrow\) BPT vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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)