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d)
\(x\ne a,x\ne b\)
đặt \(\frac{x-a}{x-b}=t\Leftrightarrow t+\frac{1}{t}=2\Leftrightarrow\frac{t^2-2t+1}{t}=0\Rightarrow t=1\)
\(\frac{x-a}{x-b}=1\Leftrightarrow\frac{\left(x-a\right)-\left(x-b\right)}{x-b}=\frac{b-a}{x-b}=0\)
Vậy: \(a\ne b\) Pt vô nghiệm
a=b phương trinhg nghiệm với mọi x khác a, b
Lời giải:
a) ĐKXĐ:
\(\left\{\begin{matrix} 2x+10\neq 0\\ x\neq 0\\ 2x(x+5)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq -5\\ x\neq 0\end{matrix}\right.\)
b)
\(B=\frac{x(x^2+2x)}{x(2x+10)}+\frac{(2x+10)(x-5)}{x(2x+10)}+\frac{50-5x}{x(2x+10)}\)
\(=\frac{x^3+2x^2+2(x^2-25)+50-5x}{x(2x+10)}=\frac{x^3+4x^2-5x}{2x(x+5)}=\frac{x^2+4x-5}{2(x+5)}=\frac{(x-1)(x+5)}{2(x+5)}=\frac{x-1}{2}\)
Để $B=0\Leftrightarrow \frac{x-1}{2}=0\Leftrightarrow x=1$ (thỏa mãn)
Để $B=\frac{1}{4}\Leftrightarrow \frac{x-1}{2}=\frac{1}{4}$
$\Leftrightarrow x-1=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}$ (thỏa mãn)
Lời giải:
a) ĐKXĐ:
\(\left\{\begin{matrix} 2x+10\neq 0\\ x\neq 0\\ 2x(x+5)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq -5\\ x\neq 0\end{matrix}\right.\)
b)
\(B=\frac{x(x^2+2x)}{x(2x+10)}+\frac{(2x+10)(x-5)}{x(2x+10)}+\frac{50-5x}{x(2x+10)}\)
\(=\frac{x^3+2x^2+2(x^2-25)+50-5x}{x(2x+10)}=\frac{x^3+4x^2-5x}{2x(x+5)}=\frac{x^2+4x-5}{2(x+5)}=\frac{(x-1)(x+5)}{2(x+5)}=\frac{x-1}{2}\)
Để $B=0\Leftrightarrow \frac{x-1}{2}=0\Leftrightarrow x=1$ (thỏa mãn)
Để $B=\frac{1}{4}\Leftrightarrow \frac{x-1}{2}=\frac{1}{4}$
$\Leftrightarrow x-1=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}$ (thỏa mãn)
\(b,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\) \(\left(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne-3\end{matrix}\right.\right)\)
\(\Leftrightarrow\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{\left(x-1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x^2+8x-3\right)-\left(2x^2+3x-5\right)+4=x^2+2x-3\)
\(\Leftrightarrow x^2+5x+6=x^2+2x-3\)
\(\Leftrightarrow9=-3x\)
\(\Leftrightarrow x=-3\left(ktmđk\right)\)
\(\Leftrightarrow Ptvn\)
\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem