1.Thay m=-1 vào pt ta được:

\(x^4-2x^2-3=0\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vn\right)\\x^2=3\end{matrix}\right.\)\(\Rightarrow x=\pm\sqrt{3}\)

Vậy...

2.Đặt \(t=x^2\left(t\ge0\right)\)

Với mỗi t>0 thì sẽ luôn có hai x phân biệt

Pttt: \(t^2-2t+m-2=0\) (2)

Để pt (1) có 4 nghiệm pb \(\Leftrightarrow\) PT (2) có hai nghiệm pb dương

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S=2>0\left(lđ\right)\\P=m-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}4-4\left(m-2\right)>0\\m>2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m< 3\\m>2\end{matrix}\right.\)\(\Rightarrow2< m< 3\)

Vậy...

1,\(\sqrt{\left(x-1\right)^2}=\left|x-1\right|=-\left(x-1\right)=1-x\)

2,\(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=-\left(a-2b\right)=2b-a\)

3,\(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)

Hoàng Khương ừ,c lười quá

a) Đk:\(x\ne0;x\ne2\)

\(A=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^2\left(x-2\right)+4\left(x-2\right)}\right].\dfrac{x^2-x-2}{x^2}\)

\(=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right].\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x+1}{2x}\)

b)Tại \(x=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\) (tm đk) thay vào A ta được:

\(A=\dfrac{\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)\(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{4}\)

Vậy...

đK: \(x\ge0;x\ne25;x\ne9\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right]:\left[\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right]\)

\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right]:\dfrac{25-x-\left(x-9\right)+\left(x-25\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\sqrt{x}-3}{\sqrt{x}+5}\)

\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)

\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)

\(=\left|4-3\sqrt{2}\right|-\sqrt{\left(3\sqrt{2}+1\right)^2}\)

\(=3\sqrt{2}-4-\left(3\sqrt{2}+1\right)\)

\(=-5\)