Question

\(1+\frac{45}{y^2-8y+16}=\frac{14}{y-4}\)

\(\frac{5}{x-1}-\frac{4}{3-6x+3x^2}=3\)

\(\frac{2x-5}{x-5}=3\)

\(\frac{x^2-12}{x}=x+\frac{3}{2}\)

\(\frac{\left(x^2-4\right)-\left(3x-6\right)}{x-2}=0\)

\(\frac{8}{2x+1}=2x-1\)

Answer 1

a/ĐKXĐ: \(y\ne4\)

Đặt \(y-4=x\)

\(1+\frac{45}{x^2}=\frac{14}{x}\Leftrightarrow x^2-14x+45=0\Rightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y-4=9\\y-4=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=13\\y=9\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1\)

Đặt \(x-1=y\)

\(\frac{5}{y}-\frac{4}{3y^2}=3\Leftrightarrow9y^2=15y-4\)

\(\Leftrightarrow9y^2-15y+4=0\Rightarrow\left[{}\begin{matrix}y=\frac{4}{3}\\y=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{4}{3}\\x-1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{4}{3}\end{matrix}\right.\)

Answer 2

c/ ĐKXĐ: \(x\ne5\)

\(\Leftrightarrow2x-5=3x-15\)

\(\Leftrightarrow x=10\)

d/ ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow2\left(x^2-12\right)=2x^2+3x\)

\(\Leftrightarrow3x=-24\Rightarrow x=-8\)

e/ ĐKXĐ: \(x\ne2\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=1\end{matrix}\right.\)

f/ DKXĐ: \(x\ne-\frac{1}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=8\)

\(\Leftrightarrow4x^2-1=8\)

\(\Leftrightarrow x^2=\frac{9}{4}\Rightarrow x=\pm\frac{3}{2}\)