Bạn thêm giúp mình dòng cuối cả hai nghiệm là thỏa mãn điều kiện nhé.

\(\dfrac{60}{x}-\dfrac{55}{21-x}=1\) (Điều kiện: \(x\ne0;x\ne21\))

\(\Leftrightarrow\dfrac{60\left(21-x\right)}{x\left(21-x\right)}-\dfrac{55x}{x\left(21-x\right)}=\dfrac{x\left(21-x\right)}{x\left(21-x\right)}\)

\(\Rightarrow1260-60x-55x=21x-x^2\)

\(\Leftrightarrow x^2-136x+1260=0\)

\(\Leftrightarrow x^2-126x-10x+1260=0\)

\(\Leftrightarrow x\left(x-126\right)-10\left(x-126\right)=0\)

\(\Leftrightarrow\left(x-126\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-126=0\\x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=126\\x=10\end{matrix}\right.\)

\(\dfrac{5}{x-3}=\dfrac{x+4}{12}\)

\(\Rightarrow\left(x-3\right)\left(x+4\right)=5.12\)

\(\Rightarrow x^2+x-12=60\)

\(\Rightarrow x^2+x-72=0\)

\(\Rightarrow x^2+9x-8x-72=0\)

\(\Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\)

\(\Rightarrow\left(x+9\right)\left(x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)

\(a,\left|2-x\right|=0\)

\(\Rightarrow2-x=0\)

\(\Rightarrow x=2\)

\(b,\left|3x-4\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}3x-4=2\\3x-4=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,\left|x-0,5\right|=-2,5\)

Vì: \(\left|x-0,5\right|\ge0\forall x\)

Mà: \(\left|x-0,5\right|=-2,5\) (vô lí)

\(\Rightarrow x\in\varnothing\)

\(d,5\left|x-0,5\right|-2=-\left|-\dfrac{2}{5}\right|\)

\(\Rightarrow5\left|x-0,5\right|-2=-0,4\)

\(\Rightarrow5\left|x-0,5\right|=1,6\)

\(\Rightarrow\left|x-0,5\right|=0,32\)

\(\Rightarrow\left[{}\begin{matrix}x-0,5=0,32\\x-0,5=-0,32\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0.82\\x=0,18\end{matrix}\right.\)