a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)
b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)
\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)
\(\Leftrightarrow2x+4-8x+24=1-x\)
\(\Leftrightarrow2x-8x+x=1-4-24\)
\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)
\(\Rightarrow S=\left\{9\right\}\)
-Mệt -.-
c. \(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=-x-2\\2x+3=x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=-1\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1;\dfrac{-5}{3}\right\}\)
d. \(\dfrac{2x-1}{3}-x-1=\dfrac{x+2}{4}\)
\(\Leftrightarrow4\left(2x-1\right)-12\left(x+1\right)=3\left(x+2\right)\)
\(\Leftrightarrow8x-4-12x-12=3x+6\)
\(\Leftrightarrow8x-12x-3x=6+4+12\)
\(\Leftrightarrow-7x=22\Leftrightarrow x=\dfrac{-22}{7}\)
\(\Rightarrow S=\left\{\dfrac{-22}{7}\right\}\)
e. \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\) (ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow x^2-4x+x-4+x^2+4x-x-4=2x^2-2\)
\(\Leftrightarrow2x^2-8=2x^2-2\Leftrightarrow0x=6\left(vl\right)\)
\(\Rightarrow S=\left\{\varnothing\right\}\)
e) \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)
ĐKXĐ : \(x\ne\pm1\) ; \(x\ne0\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=2\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=2\)
\(\Leftrightarrow\dfrac{x^2+x-4x-4+x^2-x+4x-4}{\left(x-1\right)\left(x+1\right)}=2\)
\(\Leftrightarrow\dfrac{\left(x^2+x^2\right)+\left(x-4x-x+4x\right)+\left(-4-4\right)}{\left(x-1\right)\left(x+1\right)}=2\)
\(\Leftrightarrow\dfrac{2x^2-8}{\left(x-1\right)\left(x+1\right)}=2\)
\(\Leftrightarrow\dfrac{2x^2-8}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x^2-8=2\left(x-1\right)\left(x+1\right)\)
<=> 2x2-8=(2x-2)(x+1)
<=> 2x2-8=2x2+2x-2x-2
<=> 2x2-8=2x2-2
<=> (2x2-8)-(2x2-2)=0
<=> 2x2-8-2x2+2=0
<=> -6=0 ( x vô nghiệm )
Vậy phương trình vô nghiệm
f) \(\dfrac{3x}{x-2}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
ĐKXĐ : \(x\ne2\) ; \(x\ne5\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)+3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{3x^2-15x+3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{3x^2-12x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{3x^2-12x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}\)
\(\Rightarrow3x^2-12x=x\left(x-2\right)\)
\(\Leftrightarrow\left(3x^2-12x\right)-x\left(x-2\right)=0\)
\(\Leftrightarrow3x^2-12x-x^2+2x=0\)
\(\Leftrightarrow2x^2-10x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=5\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm x=0
f. \(\dfrac{3x}{x-2}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\) (ĐKXĐ: \(x\ne2;x\ne5\))
\(\Leftrightarrow3x\left(x-5\right)+3x=x\left(x-2\right)\)
\(\Leftrightarrow3x^2-15x+3x=x^2-2x\)
\(\Leftrightarrow3x^2-x^2-12x+2x=0\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=5\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{0\right\}\)
