\(a,\dfrac{x^2-4x+4}{x-2}=1\) (1)
Đkxđ: \(x\ne2\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(x-2\right)^2}{x-2}=1\)
\(\Leftrightarrow x-2=1\Rightarrow x=3\)
\(b,\dfrac{x^2-10x+25}{x^2-25}=0\left(1\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow\dfrac{x-5}{x+5}=0\)
\(\Rightarrow x-5=0\Rightarrow x=5\)
a, ĐKXĐ:x khác 2
\(\dfrac{\left(x-2\right)^2}{x-2}\)=1
=> x - 2=1
=> x=3 ( t/m)
b,ĐKXĐ:x khác 5,-5
\(\dfrac{\left(x-5^2\right)}{\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{x-5}{x+5}\)=0
vì x+5 khác 0
=>x-5=0
=>x=5 (ko t/m)
Vậy pt vô no
a) \(\dfrac{x^2-4x+4}{x-2}=1\)
ĐKXĐ: \(x\ne2\)
\(\dfrac{\left(x-2\right)^2}{x-2}=1\)
\(x-2=1\Rightarrow x=3\)(nhận)
b) \(\dfrac{x^2-10x+25}{x^2-25}=0\)
ĐKXĐ: \(x\ne5\), \(x\ne-5\)
\(\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=0\)
\(\dfrac{x-5}{x+5}=0\)
\(x-5=0\Rightarrow x=5\)
\(x+5=0\Rightarrow x=-5\)
