a, \(\dfrac{2x}{x+2}-\dfrac{2}{x-1}=-1\)
\(\Leftrightarrow\dfrac{2x^2-2x-2x-4}{\left(x+2\right)\left(x-1\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-4x-4}{x^2-x+2x-2}=-1\)
\(\Leftrightarrow2x^2-4x-4=-x^2+x-2x+2\)
\(\Leftrightarrow3x^2+3x-6=0\)
\(\Leftrightarrow3x^2+6x-3x-6=0\)
\(\Leftrightarrow3x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy...
b, \(\dfrac{x+3}{x-4}+\dfrac{x-1}{x-2}=\dfrac{2}{6x-x^2-8}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}=\dfrac{2}{6x-x^2-8}\)
\(\Leftrightarrow\dfrac{x^2-2x+3x-6+x^2-4x-x+4}{x^2-2x-4x+8}=\dfrac{2}{6x-x^2-8}\)
\(\Leftrightarrow\dfrac{2x^2-4x-2}{x^2-6x+8}=\dfrac{2}{6x-x^2-8}\)
\(\Leftrightarrow2x^2-4x-2=-2\)
\(\Leftrightarrow x^2-2x-1=-1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy...
c, \(\left|-3x\right|=x-7\)
\(\Leftrightarrow\left|3x\right|=x-7\)
+) Xét \(x\ge0\) có:
\(3x=x-7\)
\(\Leftrightarrow2x=-7\) ( không t/m )
+) Xét x < 0 có:
\(-3x=x-7\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{7}{4}\) ( ko t/m )
Vậy không có giá trị x thỏa mãn
d, +) Xét \(x\ge4\) có:
\(3x+x-4=0\)
\(\Leftrightarrow4x=4\Leftrightarrow x=1\) ( ko t/m )
+) Xét \(0\le x< 4\) có:
\(3x+4-x=0\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) ( ko t/m )
+) Xét \(x< 0\) có:
\(-3x+4-x=0\)
\(\Leftrightarrow-4x=-4\)
\(\Leftrightarrow x=1\) ( ko t/m )
Vậy không có giá trị x thỏa mãn
e, tương tự phần c
