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8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
1. \(\Leftrightarrow\left(3x-1\right)\left(\sqrt{5}x-2\right)\ge0\Rightarrow\left[{}\begin{matrix}x\le\frac{1}{3}\\x\ge\frac{2}{\sqrt{5}}\end{matrix}\right.\)
2. \(\Leftrightarrow\frac{\left(3-2x\right)\left(3+2x\right)}{2x-3}\ge0\Leftrightarrow\left[{}\begin{matrix}x\ne\frac{3}{2}\\x\le-\frac{3}{2}\end{matrix}\right.\)
3. \(\left|x-2\right|\ge3\Leftrightarrow\left[{}\begin{matrix}x-2\ge3\\x-2\le-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)
4. \(\Leftrightarrow-10\le3x+1\le10\Rightarrow-\frac{11}{3}\le x\le3\)
5. \(\Leftrightarrow\frac{3x^2-x+2}{x^2-9}-3\le0\Leftrightarrow\frac{-x+29}{\left(x-3\right)\left(x+3\right)}\le0\Rightarrow\left[{}\begin{matrix}-3< x< 3\\x\ge29\end{matrix}\right.\)
6. \(\Leftrightarrow\frac{4}{\left(x-2\right)^2}+\frac{1}{x-2}>0\Leftrightarrow\frac{x+2}{\left(x-2\right)^2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-2\\x\ne2\end{matrix}\right.\)
a/ \(\Leftrightarrow\frac{3x+4}{x\left(x+4\right)}-\frac{3}{x+3}\le0\)
\(\Leftrightarrow\frac{\left(3x+4\right)\left(x+3\right)-3x\left(x+4\right)}{x\left(x+3\right)\left(x+4\right)}\le0\)
\(\Leftrightarrow\frac{x+12}{x\left(x+3\right)\left(x+4\right)}\le0\)
\(\Rightarrow\left[{}\begin{matrix}-12\le x< -4\\-3< x< 0\end{matrix}\right.\)
b/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le-\frac{3}{4}\end{matrix}\right.\)
- Với \(x< \frac{1}{2}\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) \(\Rightarrow BPT\) vô nghiệm
- Với \(x\ge\frac{1}{2}\Rightarrow x\ge4\), hai vế đều không âm, bình phương 2 vế:
\(\Leftrightarrow4x^2-13x-12< \left(2x-1\right)^2\)
\(\Leftrightarrow4x^2-13x-12< 4x^2-4x+1\)
\(\Leftrightarrow9x>-13\Rightarrow x>-\frac{13}{9}\)
Kết hợp điều kiện ta được \(x\ge4\)
Vậy nghiệm của BPT là \(x\ge4\)