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\(ĐKXĐ:\hept{\begin{cases}x^2-8x+15\ge0\\x^2+2x-15\ge0\\4x^2-18x+18\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le-5\\x=3\end{cases}}\)
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 \(x\in(-\infty;-5]\mu\left\{3\right\}\mu\left[5;\frac{17}{3}\right]\)
Câu 1:
Xét \(m=0\Rightarrow f\left(x\right)=0-0-1\le0\left(lđ\right)\)
Xét \(m>0\Rightarrow f\left(x\right)\le0\Leftrightarrow x_1\le0< 3\le x_2\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\f\left(3\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le0\left(lđ\right)\\9m-6m-1\le0\end{matrix}\right.\Leftrightarrow m\le\frac{1}{3}\Rightarrow0< m\le\frac{1}{3}\)
Xét \(m< 0\Rightarrow f\left(x\right)\le0\)
Chia làm 3 TH:
TH1: \(\Delta< 0\Leftrightarrow m\left(m+1\right)< 0\Leftrightarrow-1< m< 0\)
TH2: \(\Delta=0\Rightarrow m\left(m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=-1\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}0\le x_1< x_2\\x_1< x_2\le3\end{matrix}\right.\end{matrix}\right.\)
\(\Delta>0\Leftrightarrow m< -1\)
\(0\le x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\\frac{x_1+x_2}{2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le0\left(lđ\right)\\\frac{2m}{m}>0\left(lđ\right)\end{matrix}\right.\)
\(x_1< x_2\le3\Leftrightarrow\left\{{}\begin{matrix}f\left(3\right)\le0\\\frac{x_1+x_2}{2}< 3\left(lđ\right)\end{matrix}\right.\)
Vậy \(m\in\left[-1;\frac{1}{3}\right]\)
Có gì sai sót bảo mình ạ :<
ĐKXĐ: \(x\ge\frac{2}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x+1}=a>0\\\sqrt{3x-2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x+3\)
Phương trình trở thành:
\(a-b=\frac{a^2-b^2}{5}\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-5\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{4x+1}=\sqrt{3x-2}\left(1\right)\\\sqrt{4x+1}+\sqrt{3x-2}=5\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x+1=3x-2\Rightarrow x=-3< \frac{2}{3}\left(l\right)\)
\(\left(2\right)\Leftrightarrow4x+1+3x-2+2\sqrt{\left(4x+1\right)\left(3x-2\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(4x+1\right)\left(3x-2\right)}=26-7x\) (\(\frac{2}{3}\le x\le\frac{26}{7}\))
\(\Leftrightarrow4\left(4x+1\right)\left(3x-2\right)=\left(26-7x\right)^2\)
\(\Leftrightarrow...\)