Bài `1`

\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\\ =\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\\ =\left|\sqrt{3}-\sqrt{5}\right|-\left|\sqrt{3}+\sqrt{5}\right|\\ =\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =-2\sqrt{3}\)

\(B=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\\ B-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot2+1^2}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)

Mà \(A-\sqrt{2}=-2\)

\(\Rightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)

2) \(C=x-\sqrt{x-3}\)

\(C=\left(x-3\right)-\sqrt{x-3}+3\)

Đặt \(\sqrt{x-3}=t\left(t\ge0\right)\)

\(C=t^2-t+3\)

\(C=t^2-2t.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}\)

\(C=\left(t-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\) \(\ge\dfrac{11}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow t=\dfrac{1}{2}\Leftrightarrow\sqrt{x-3}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{13}{4}\)

Vậy GTNN của C là \(\dfrac{11}{4}\)

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{x-1}{\sqrt{x}}\)

https://edward29.github.io/surprise/

** Sửa đề: $m\neq 0; m\neq -1$

Lời giải:

Gọi đths đã cho là $(d)$.

Gọi $A,B$ lần lượt là giao điểm của $(d)$với trục $Ox, Oy$.

Do $A\in Ox$ nên $y_A=0$

$A\in (d)\Rightarrow y_A=mx_A+x_A+1$

$\Leftrightarrow 0=x_A(m+1)+1$

$\Leftrightarrow x_A=\frac{-1}{m+1}$

Do $B\in Oy$ nên $x_B=0$

$y_B=mx_B+x_B+1=m.0+0+1=1$

Gọi $h$ là khoảng cách từ gốc tọa độ đến $(d)$. 

Theo hệ thức lượng trong tam giác vuông:

$\frac{1}{h^2}=\frac{1}{OA^2}+\frac{1}{OB^2}$

$\Leftrightarrow \frac{1}{h^2}=\frac{1}{x_A^2}+\frac{1}{y_B^2}$

$\Leftrightarrow \frac{1}{h^2}=1+(m+1)^2$

Với $m\neq -1$ thì không tìm được min $1+\frac{1}{(m+1)^2}$, tức là không tìm được max h. 

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