a) \(y=m\left(2x-1\right)+3-2x,\forall m\)

\(\Leftrightarrow m\left(2x-1\right)+3-2x-y=0,\forall m\)

\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\3-2x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=2\end{cases}}\)

Vậy khi \(m\)thay đổi đường thẳng \(\left(d\right)\)luôn đi qua điểm có tọa độ \(\left(\frac{1}{2},2\right)\).

b) \(y=m\left(2x-1\right)+3-2x=\left(2m-2\right)x+3-m\)

\(\Leftrightarrow y-\left(2m-2\right)x+m-3=0\)

Khoảng cách từ điểm \(O\left(0,0\right)\)đến đường thẳng \(d\)là: 

\(d=\frac{\left|m-3\right|}{\sqrt{\left(2m-2\right)^2+1^2}}\Leftrightarrow d^2\left(4m^2-8m+5\right)=m^2-6m+9\)

\(\Leftrightarrow m^2\left(4d^2-1\right)-2m\left(4d^2-3\right)+5d^2-9=0\)(1)

Với \(m=0\): \(d=\frac{3\sqrt{5}}{5}\).

Với \(m\ne0\)ta coi \(m\)là phương trình bậc \(2\)ẩn \(m\)tham số \(d\).

Để phương trình có nghiệm thì 

\(\Delta'\ge0\Leftrightarrow\left(4d^2-3\right)^2-\left(5d^2-9\right)\left(4d^2-1\right)\ge0\)

\(\Leftrightarrow17d^2-4d^4\ge0\)

\(\Leftrightarrow\frac{-\sqrt{17}}{2}\le d\le\frac{\sqrt{17}}{2}\).

Vây GTLN cần tìm là \(d=\frac{\sqrt{17}}{2}\).

ĐKXĐ : \(x\ge\frac{1}{2}\)

=> \(2+\sqrt{2x-1}=x\)

<=> \(\sqrt{2x-1}=x-2\)

<=> \(\left(\sqrt{2x-1}\right)^2=\left(x-2\right)^2\)

<=> 2x - 1 = x² - 4x + 4

<=> x² - 6x + 5 = 0

<=> (x - 1)(x - 5) = 0

<=> \(\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

Xyz thay x = 1 vô thì không xảy ra đẳng thức đâu

ĐK : x >= 2 còn lại làm như bạn

ta có :

\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}\)

\(=8-2\sqrt{16-7}=8-2\sqrt{9}=2\)

vậy \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}\)

\(\sqrt{4-\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)=\left(4+\sqrt{15}\right)\sqrt{8-2\sqrt{15}.}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2>\sqrt{3}\)

vậy số bên trái lơn hơn \(\sqrt{3}\)

\(\sqrt{11-4\sqrt{7}}-\sqrt{29-4\sqrt{7}}\)

\(=\sqrt{7-2.\sqrt{7}.2+4}-\sqrt{28-2.2\sqrt{7}.1+1}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}-\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(2\sqrt{7}-1\right)^2}\)

\(=\left|\sqrt{7}-2\right|-\left|2\sqrt{7}-1\right|\)

\(=\sqrt{7}-2-2\sqrt{7}+1\)

\(=-\sqrt{7}-1\)

Bài 2. 

ĐKXĐ của biểu thức đã cho là: 

\(\hept{\begin{cases}x\ge0,\sqrt{x}\ne0\\\sqrt{x}-1\ne0\\\sqrt{x}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1,x\ne2\end{cases}}\).

\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(A>\frac{1}{6}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}>\frac{1}{6}\Leftrightarrow6\left(\sqrt{x}-2\right)>\sqrt{x}\)

\(\Leftrightarrow5\sqrt{x}>12\Leftrightarrow x>\frac{144}{25}\).

ĐK: \(-\sqrt{3}\le x\le\sqrt{3}\).

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

\(\Leftrightarrow\left(x-2\right)\sqrt{3-x^2}=\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(\sqrt{3-x^2}-x-1\right)=0\)

\(\Leftrightarrow\sqrt{3-x^2}=x+1\)(vì \(-\sqrt{3}\le x\le\sqrt{3}\))

\(\Rightarrow3-x^2=\left(x+1\right)^2\)

\(\Leftrightarrow2x^2+2x-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\left(tm\right)\\x=\frac{-1-\sqrt{5}}{2}\left(l\right)\end{cases}}\)