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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+61x\ge0\\4x+2\ge0\\x^2+61x\le\left(4x+2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge0\\x\le-61\end{matrix}\right.\\x\ge-\dfrac{1}{2}\\15x^2-45x+4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge\dfrac{45+\sqrt{1785}}{30}\\x\le\dfrac{45-\sqrt{1785}}{30}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0\le x\le\dfrac{45-\sqrt{1785}}{30}\\x\ge\dfrac{45+\sqrt{1785}}{30}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(3-x\right)\left(3+x\right)}>0\)
Bảng xét dấu:
Từ bảng xét dấu ta có nghiệm của BPT là: \(x\in\left(-3;1\right)\cup\left(2;3\right)\)
1) ĐK: \(x\ge-1\)
\(\sqrt{9x^2+9x+4}>9x+3-\sqrt{x+1}\)
<=> \(\sqrt{9x^2+9x+4}+\sqrt{x+1}>9x+3\)(1)
TH1: 9x + 3 \(\le\)0 <=> x\(\le-\frac{1}{3}\)
(1) luôn đúng
Th2: x\(>-\frac{1}{3}\)
<=> \(\left(\frac{1}{2}x+1-\sqrt{x+1}\right)+\left(\frac{17}{2}x+2-\sqrt{9x^2+9x+4}\right)< 0\)
<=> \(\frac{\frac{1}{4}x^2}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{\frac{253}{4}x^2}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}< 0\)
<=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)< 0\)vô nghiệm
Vì với x \(>-\frac{1}{3}\):
ta có: \(\frac{1}{2}x+1+\sqrt{x+1}>0\)
\(\frac{17}{2}x+2+\sqrt{9x^2+9x+4}=\frac{17}{2}x+2+\sqrt{3\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}>\frac{17}{2}x+2+1>0\)
=> \(\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)>0\)với x \(>-\frac{1}{3}\) và \(x^2\ge0\)với mọi x
=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)\ge0\)với x\(>-\frac{1}{3}\)
Vậy \(x< -\frac{1}{3}\)
Xin lỗi bạn kết luận bài 1 là:
\(-1\le x\le-\frac{1}{3}\)
Bài 2) \(2+\sqrt{x+2}-x\sqrt{x+2}=x\left(\sqrt{x+2}-x\right)\)(2)
ĐK: \(x\ge-2\)
(2) <=> \(2+\sqrt{x+2}+x^2-2x\sqrt{x+2}=0\)
<=> \(8+4\sqrt{x+2}+4x^2-8x\sqrt{x+2}=0\)
<=> \(\left(2x-1\right)^2-4\left(2x-1\right)\sqrt{x+2}+4\left(x+2\right)-1=0\)
<=> \(\left(2x-1-2\sqrt{x+2}\right)^2-1=0\)
<=> \(\left(x-1-\sqrt{x+2}\right)\left(x-\sqrt{x+2}\right)=0\)
<=> \(\orbr{\begin{cases}x-1=\sqrt{x+2}\left(3\right)\\x=\sqrt{x+2}\left(4\right)\end{cases}}\)
(3) <=> \(\hept{\begin{cases}x\ge1\\x^2-3x-1=0\end{cases}}\Leftrightarrow x=\frac{3+\sqrt{13}}{2}\left(tm\right)\)
(4) <=> \(\hept{\begin{cases}x\ge0\\x^2-x-2=0\end{cases}\Leftrightarrow}x=2\left(tm\right)\)
Kết luận:...
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
\(\Leftrightarrow\left|x^2-9x+14\right|>x^2-3x-4\)
Trường hợp 1: \(\left\{{}\begin{matrix}x^2-9x+14>0\\x^2-3x-4< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left(-\infty;2\right)\cup\left(7;+\infty\right)\\-1< =x< =4\end{matrix}\right.\)
\(\Leftrightarrow x\in[-1;2)\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x^2-3x-4>=0\\\left(x^2-9x+14-x^2+3x+4\right)\left(x^2-9x+14+x^2-3x-4\right)>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+1\right)>=0\\\left(-6x+18\right)\left(2x^2-12x+10\right)>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in[-\infty;-1)\cup[4;+\infty)\\\left(x-3\right)\left(x^2-6x+5\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in[-\infty;-1)\cup[4;+\infty)\\x\in[-\infty;1]\cup\left(3;5\right)\end{matrix}\right.\Leftrightarrow x\in[-\infty;-1)\cup[4;5)\)