\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
\(=x-3\)
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x+4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
= x - 3
cho mình làm lại
ĐKXĐ:x khác 4
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x-4\right|+\frac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x-4\right|+\frac{x-4}{\left|x-4\right|}\)
+)Nếu x<4
PT trở thành \(4-x+\frac{x-4}{4-x}=4-x+\left(-1\right)=3-x\)
+)Nếu x > hoặc = 4
PT trở thành \(x-4+\frac{x-4}{x-4}=x-4+1=x-3\)
Vậy ...
