1) Với m = 1 thì ta có:

\(x^2-2\left(1-1\right)x+2\cdot1-3=0\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

2) Ta có: \(\Delta^'=\left[-\left(m-1\right)\right]^2-\left(2m-3\right)\cdot1=m^2-2m+1-2m+3\)

\(=m^2-4m+4=\left(m-2\right)^2\ge0\left(\forall m\right)\)

=> PT luôn có nghiệm với mọi m

Theo hệ thức viet ta có:

\(\hept{\begin{cases}x_1+x_2=2\left(m-1\right)\\x_1x_2=2m-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x_1+x_2-1=2m-3\\x_1x_2=2m-3\end{cases}}\)

\(\Rightarrow x_1+x_2-1=x_1x_2\)

\(\Leftrightarrow x_1x_2-\left(x_1+x_2\right)+1=0\)

\(\left(\sqrt{45}-\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(\sqrt{9.5}-\sqrt{9.7}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=-3\left(\sqrt{5}-\sqrt{7}\right)^2\)

Ta có:

\(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}\cdot\frac{1+y}{8}\cdot\frac{1+z}{8}}=\frac{3}{4}x\)

Tương tự:

\(\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge\frac{3}{4}y\)

\(\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge\frac{3}{4}z\)

\(\Rightarrow VT+\frac{3}{4}+\frac{1}{4}\left(x+y+z\right)\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Leftrightarrow VT\ge\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{1}{2}\cdot3\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi: x = y = z = 1

Ta có : \(B=\frac{2\sqrt{x}-x}{\sqrt{x}-2}=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-2}\)

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

Mình chưa vẽ hình nhưng mà câu c bạn có sai không? Tại vì bạn ghi thế thì có khác gì chứng minh AK=AD đâu. Bạn xem lại nhá 

\(\frac{2}{AK}=\frac{1}{AD}+\frac{1}{AE}\) nhá