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a/ \(\frac{3}{1-x}-\frac{5}{2x+1}\ge0\Leftrightarrow\frac{11x-2}{\left(1-x\right)\left(2x+1\right)}\ge0\Rightarrow\left[{}\begin{matrix}x< -\frac{1}{2}\\\frac{2}{11}\le x< 1\end{matrix}\right.\)
b/ \(\frac{\left(2x-1\right)\left(2-x\right)}{\left(x-1\right)\left(x-3\right)}< 0\Rightarrow\left[{}\begin{matrix}x>3\\1< x< 2\\x< \frac{1}{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
b/ ĐKXĐ: \(x\ne\left\{-\frac{4}{3};1\right\}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow44x=-33\Rightarrow x=-\frac{3}{4}\)
c/ ĐKXĐ: \(x\ne\left\{-\frac{1}{4};0\right\}\)
\(\Leftrightarrow\frac{3\left(x^2-1\right)}{4x+1}+\frac{2\left(1-x^2\right)}{x}-\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{3}{4x+1}-\frac{2}{x}-1\right)=0\)
TH1: \(x^2-1=0\Rightarrow x=\pm1\)
TH2: \(\frac{3}{4x+1}-\frac{2}{x}-1=0\Leftrightarrow3x-2\left(4x+1\right)-x\left(4x+1\right)=0\)
\(\Leftrightarrow4x^2+6x+2=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)
a/ ĐKXĐ:...
\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-1\right)^2}+2\left|\frac{x-2}{x-1}\right|-3=0\)
\(\Leftrightarrow\left|\frac{x-2}{x-1}\right|^2+2\left|\frac{x-2}{x-1}\right|-3=0\)
Đặt \(\left|\frac{x-2}{x-1}\right|=t\left(t\ge0\right)\)
\(\Rightarrow pt\Leftrightarrow t^2+2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
Ok tự giải nốt
b/ viết lại đề bài đi cậu
a/ \(\Leftrightarrow\frac{2x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\le0\)
\(\Leftrightarrow\frac{5\left(-3x+4\right)}{\left(x-5\right)\left(x+5\right)}\le0\) \(\Rightarrow\left[{}\begin{matrix}-5< x\le\frac{4}{3}\\x>5\end{matrix}\right.\)
b/ Không rõ đề
c/
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT>0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm
- Với \(x\ge-1\) hai vế ko âm, bình phương:
\(\Leftrightarrow\left(x+1\right)^2>\frac{\left(x-3\right)^2}{4}\)
\(\Leftrightarrow\left(2x+2\right)^2-\left(x-3\right)^2>0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-1\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\x>\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow x>\frac{1}{3}\)