\(a)\)\(A=\sqrt{3\left(\sqrt{2}\right)^2-2\left(\sqrt{2}\right)-\sqrt{2}.\sqrt{2}-1}=\sqrt{6-2\sqrt{2}-2-1}\)

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

\(b)\)\(C=\left(1-\sqrt{2}\right)^2+\sqrt{8}-2=1-2\sqrt{2}+2+2\sqrt{2}-2=1\)

\(c_1)\)\(D=\sqrt{\left(1-\sqrt{x}\right)^2}.\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}\) ( đề sai r mem :3 ) 

\(D=\left|\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right|=\left|x-1\right|\)

\(c_2)\)\(D=\left|x-1\right|=\left|2020-1\right|=2019\)

\(x^2-4x-m^2=0\) (1) 

\(a)\) Để pt (1) có hai nghiệm phân biệt \(x_1,x_2\) thì \(\Delta'=\left(-2\right)^2-\left(-m\right)^2=4+m^2>0\) ( luôn đúng ) 

Vậy pt (1) luôn có hai nghiệm phân biệt \(x_1,x_2\) với mọi m

\(b)\) Ta có : \(A=\left|x_1^2-x_2^2\right|=\left|\left(x_1+x_2\right)\left(x_1-x_2\right)\right|\)

\(\Leftrightarrow\)\(A^2=\left(x_1+x_2\right)^2\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2\left(x_1^2+x_2^2-2x_1x_2\right)=\left(x_1+x_2\right)^2\left[\left(x_1+x_2\right)^2-4x_1x_2\right]\) (*)

Theo định lý Vi-et ta có : \(\hept{\begin{cases}x_1+x_2=4\\x_1x_2=-m^2\end{cases}}\)

(*) \(\Leftrightarrow\)\(A^2=4^2\left[4^2-4\left(-m^2\right)\right]=16\left(16+4m^2\right)=64m^2+256\ge256\)

\(\Leftrightarrow\)\(A\ge\sqrt{256}=16\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(64m^2=0\)\(\Leftrightarrow\)\(m=0\)

Vậy GTNN của \(A=16\) khi \(m=0\)

a,\(x^2-4x-m^2=0\)(*)

\(\Delta=4^2-4\left(-m^2\right)=16+4m^2\ge16>0\)

Vậy pt luôn có 2 nghiệm phân biệt với mọi giá trị của m.

b,\(x_1=\frac{4-\sqrt{4m^2+16}}{2};x_2=\frac{4+\sqrt{4m^2+16}}{2}\)

\(\Rightarrow\left|x_1+x_2\right|=\left|\frac{4-\sqrt{4m^2+16}+4+\sqrt{4m^2+16}}{2}\right|=\left|\frac{8}{2}\right|=4\)

pt luôn = 4

Sửa câu b

\(A=\left|x_1^2-x_2^2\right|=\left|\left(x_1-x_2\right)\left(x_1+x_2\right)\right|=\left|\left(\frac{4-\sqrt{4m^2+16}}{2}-\frac{4+\sqrt{4m^2+16}}{2}\right)\left(\frac{4-\sqrt{4m^2+16}}{2}+\frac{4+\sqrt{4m^2+16}}{2}\right)\right|\)\(\Leftrightarrow A=\left|-\left(\sqrt{4m^2+16}\right).4\right|\)

Vì \(4m^2+16>0\)

\(\Rightarrow A=\sqrt{4m^2+16}.4\ge\sqrt{16}.4=4^2=16\)

Vậy MinA = 16

a,\(A=\sqrt{3x^2-2x-x\sqrt{2}-1}\)

Thay \(x=\sqrt{2}\)

\(\Rightarrow A=\sqrt{3\sqrt{2}^2-2\sqrt{2}-\sqrt{2}\sqrt{2}-1}=\sqrt{6-2\sqrt{2}-2-1}=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{2}-1\)

b,\(C=\left(1-\sqrt{2}\right)^2+\sqrt{8}-2\)

\(C=1-2\sqrt{2}+2+2\sqrt{2}-2\)

\(C=1\)