\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)

\(=x-4+\frac{x-4}{\sqrt{\left(x-4\right)^2}}\)

\(=x-4+\frac{x-4}{x-4}\)

\(=x-4+1\)

\(=x-3\)

\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)

\(=x-4+\frac{x-4}{\sqrt{\left(x+4\right)^2}}\)

\(=x-4+\frac{x-4}{x-4}\)

\(=x-4+1\)

= 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 ...

Dề sai ko bạn

Chỉ cần ý b thôi 

mình xin lỗi sửa x-1 thành x+1

\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}=\) \(\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)

                                           \(=\frac{x+6\sqrt{x}+9}{3+\sqrt{x}}\)

                                            \(=\frac{\left(3+\sqrt{x}\right)^2}{3+\sqrt{x}}\)

                                             \(=3+\sqrt{x}\)

\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}\left(x\ge0\right)=\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)

\(=\frac{x+\sqrt{6}+9}{3+\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)^2}{3+\sqrt{x}}=3+\sqrt{x}\left(x\ge0\right)\)

\(\frac{1-2x-2x^2}{1-x}\)

a, Với x khác 1 

\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)

b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

Vậy với x khác 1 thì bth A luôn nhận gtri âm