Gọi \(B\left(x_0;y_0\right)\in\left(d\right)\Rightarrow y_0=-x_0+4\)

\(AB=\sqrt{\left(x_0-1\right)^2+\left(y_0-4\right)^2}\\ \Leftrightarrow AB^2=\left(x_0-1\right)^2+\left(-x_0+4-4\right)^2\\ =2x^2_0-2x_0+1=\left(\sqrt{2}x-\dfrac{1}{\sqrt{2}}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

Dễ thấy AB nhỏ nhất khi \(\left(\sqrt{2}x_0-\dfrac{1}{\sqrt{2}}\right)^2=0\Rightarrow\sqrt{2}x_0-\dfrac{1}{\sqrt{2}}=0\\ \Rightarrow x_0=\dfrac{1}{\sqrt{2}}:\sqrt{2}=\dfrac{1}{2}\Rightarrow y_0=\dfrac{7}{2}\)

Vậy \(B\left(\dfrac{1}{2};\dfrac{7}{2}\right)\) thì AB bé nhất và bằng \(\dfrac{\sqrt{2}}{2}\)

`\sqrt{x^2}-3x+2`

`=|x|-3x+2`

`@TH1:`\(x \ge 0=>|x|=x\)

  `=>x-3x+2=-2x+2`

`@TH2:x < 0=>|x|=-x`

   `=>-x-3x+2=-4x+2`

x² + \(\dfrac{18x}{5}\) - 64 = 0

△ = (18/5)² -4.(-64) = \(\dfrac{6724}{25}\)

x = { -(18/5) + - (82/5)}: 2

x ϵ {32/5; -10}

`\sqrt{64x+64}-\sqrt{25x+25}+\sqrt{4x+4}=20`     `ĐK: x >= -1`

`<=>8\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=20`

`<=>5\sqrt{x+1}=20`

`<=>\sqrt{x+1}=4`

`<=>x+1=16`

`<=>x=15` (t/m)

Với \(x \ge 0,x \ne 1\) có:

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(A=\dfrac{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2-3\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}\)

\(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}\)

\(A=\dfrac{2x-3\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}\)

\(A=\dfrac{(\sqrt{x}-1)(2\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}\)

\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

Em nhập lại nội dung câu hỏi nha em!

Tìm điều kiện để biểu thức có nghĩa nha mọi người 

Đặt \(A=\sqrt{x^2-9}+\dfrac{1}{\sqrt{x^2-9}}\)

A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\dfrac{1}{\sqrt{x^2-9}}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2\ge9\\x^2>9\end{matrix}\right.\)\(\Leftrightarrow x^2>9\)

\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\)

Chúc học tốt nhé :))