a. \(\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\) (ĐKXĐ: \(x\ne1\))
\(\Leftrightarrow2x-1+x-1=1\)
\(\Leftrightarrow3x-2=1\Leftrightarrow x=1\)(KTM)
\(\Rightarrow S=\left\{\varnothing\right\}\)
b. \(\dfrac{5x}{2x+2}+1=\dfrac{-6}{x+1}\) (ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow5x+2\left(x+1\right)=-12\)
\(\Leftrightarrow5x+2x+2=-12\Leftrightarrow7x=-14\Leftrightarrow x=-2\) (TM)
\(\Rightarrow S=\left\{-2\right\}\)
c. \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (ĐKXĐ: \(x\ne0\))
\(\Leftrightarrow x^3+x=x^4+1\Leftrightarrow-\left(x^3+x\right)=-\left(x^4+1\right)\)
\(\Leftrightarrow-x^3-x=-x^4-1\Leftrightarrow x^4-x^3-x+1=0\)
\(\Leftrightarrow x^3\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) (TM)
\(\Rightarrow S=\left\{1\right\}\)
a. 2x−1x−1+1=1x−1 (ĐKXĐ: x≠1)
⇔2x−1+x−1=1
⇔3x−2=1⇔x=1(không thỏa mãn )
⇒S={∅}
b. 5x2x+2+1=−6x+1 (ĐKXĐ: x≠−1)
⇔5x+2(x+1)=−12
⇔5x+2x+2=−12⇔7x=−14⇔x=−2 (thỏa mãn )
⇒S={−2}
c. x+1x=x2+1x2 (ĐKXĐ: x≠0)
⇔x3+x=x4+1⇔−(x3+x)=−(x4+1)
⇔−x3−x=−x4−1⇔x4−x3−x+1=0
⇔x3(x−1)−(x−1)=0⇔(x−1)2(x2+x+1)=0
⇔(x−1)2=0⇔x−1=0⇔x=1 (thỏa mãn )
