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1.ĐK: \(x\ge\dfrac{1}{4}\)
bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)
\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)
\(\Leftrightarrow20x^2-x-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)
2.ĐK: \(-2\le x\le\dfrac{5}{2}\)
bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)
\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)
\(\Leftrightarrow x^2< -x^2+x+6\)
\(\Leftrightarrow-2x^2+x+6>0\)
\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)
3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)
.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)
*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)
*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
a.
\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1\le x\le3\)
b.
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)
Lời giải:
a) ĐK: $x\geq 0$
BPT $\Leftrightarrow \sqrt{x+2}(\sqrt{2}-1)\leq \sqrt{x}$
$\Leftrightarrow (3-2\sqrt{2})(x+2)\leq x$
$\Leftrightarrow x(2-2\sqrt{2})\leq 2(2\sqrt{2}-3)$
$\Leftrightarrow x\geq \frac{2(2\sqrt{2}-3)}{2-2\sqrt{2}}=-1+\sqrt{2}$
Vậy BPT có nghiệm $x\geq -1+\sqrt{2}$
b) ĐK: $x\geq 2$ hoặc $x\leq -2$
BPT $\Leftrightarrow (x-5)\sqrt{x^2-4}-(x-5)(x+5)\leq 0$
$\Leftrightarrow (x-5)[\sqrt{x^2-4}-(x+5)]\leq 0$Ta có 2 TH:
TH1:
\(\left\{\begin{matrix} x-5\geq 0\\ \sqrt{x^2-4}-(x+5)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ \sqrt{x^2-4}\leq x+5\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ x^2-4\leq x^2+10x+25\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ 29\leq 10x\end{matrix}\right.\Leftrightarrow x\geq 5\)
TH2:
\(\left\{\begin{matrix} x-5\leq 0\\ \sqrt{x^2-4}-(x+5)\geq 0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 5\\ x^2-4\geq x^2+10x+25 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ -29\geq 10x\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ x\leq \frac{-29}{10}\end{matrix}\right.\Leftrightarrow x\leq \frac{-29}{10}\)
Kết hợp đkxđ suy ra $x\geq 5$ hoặc $x\leq \frac{-29}{10}$
Em 2k8 k biết làm có đúng k
ĐKXĐ : \(\left[{}\begin{matrix}x\le-1\\x\ge3\end{matrix}\right.\)
Bpt \(\Leftrightarrow\left(x-2\right)\left[x+2-\sqrt{x^2-2x-3}\right]\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2;x+2\ge\sqrt{x^2-2x-3}\left(1\right)\\x\le2;x+2\le\sqrt{x^2-2x-3}\left(2\right)\end{matrix}\right.\)
(1) có : \(x+2\ge\sqrt{x^2-2x-3}\Leftrightarrow\left(x+2\right)^2\ge x^2-2x-3\)
\(\Leftrightarrow6x+7\ge0\) (Đ với \(x\ge2\) )
(2) có : \(\sqrt{x^2-2x-3}\ge x+2\)
TH1 : x + 2 < 0 <=> \(x< -2\) ( Bpt luôn đúng )
TH2 : \(x+2\ge0\) ; Bp 2 vế rút gọn được : \(6x+7\le0\Leftrightarrow x\le\dfrac{-7}{6}\)
Khi đó : \(-2\le x\le\dfrac{-7}{6}\)
Vậy ...
\(ĐKXĐ:\hept{\begin{cases}x^2-8x+15\ge0\\x^2+2x-15\ge0\\4x^2-18x+18\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le-5\\x=3\end{cases}}\)
Với x = 8 thì (*) thỏa mãn \(\Rightarrow x=3\)là 1 nghiệm của bất phương trình.
\(\left(^∗\right)\Leftrightarrow\sqrt{\left(x-5\right)\left(x-3\right)}+\sqrt{\left(x+5\right)\left(x-3\right)}\le\sqrt{\left(x-3\right)\left(4x-6\right)}\)(1)
Với \(x\ge5\Rightarrow x-3\ge2>0\)hay \(x-3>0\)thì
\(\left(1\right)\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}\le\sqrt{4x-6}\)\(\Leftrightarrow2x+2\sqrt{x^2-25}\le4x-6\)
\(\Leftrightarrow\sqrt{x^2-25}\le x-3\Leftrightarrow x^2-25=x^2-6x+9\Leftrightarrow x\le\frac{17}{3}\)
\(\Rightarrow5\le x\le\frac{17}{3}\)
Với \(x\le-5\Leftrightarrow-x\ge5\Leftrightarrow3-x\ge8>0\)hay \(x\le-5\Leftrightarrow-x\ge5\Leftrightarrow3-x>0\)thì
\(\left(1\right)\Leftrightarrow\sqrt{\left(5-x\right)\left(3-x\right)}+\sqrt{\left(-5-x\right)\left(3-x\right)}\)
\(\le\sqrt{\left(3-x\right)\left(4-6x\right)}\)
\(\Leftrightarrow\sqrt{5-x}+\sqrt{-x-5}\le\sqrt{6-4x}\)
\(\Leftrightarrow-2x+2\sqrt{\left(5-x\right)\left(-x-5\right)}\le6-4x\)
\(\Leftrightarrow\sqrt{x^2-25}\le3-x\Leftrightarrow x^2-25\le x^2-6x+9\)
\(\Leftrightarrow x\le\frac{17}{3}\Rightarrow x\le-5\)
Từ đó suy ra tập nghiệm của bpt là \(x\in(-\infty;-5]\mu\left\{3\right\}\mu\left[5;\frac{17}{3}\right]\)