AI=căn 4^2-2,4^2=3,2cm

AB=4^2/3,2=5cm

BH=căn 5^2-4^2=3cm

BC=AB^2/BH=25/3(cm)

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

ĐKXĐ: -x^2+5x-4>=0 và 2x-7<>0

=>1<=x<=4 và x<>7/2

a) \(C>0\) Khi:

\(\dfrac{-2}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Vậy: ..

b) \(C< 0\) khi:

\(\dfrac{-2}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow\sqrt{x}>3\)

\(\Leftrightarrow x>9\)

Vậy: ...

`a, C > 0 <=> sqrt x - 3 < 0`.

`<=> sqrt x < 3`

`<=> x < 9`.

`b, C < 0 <=> sqrt x - 3 > 0`

`<=> sqrt x > 3`

`<=> x > 9`

\(\sqrt{-x^2+5x-4}+\dfrac{1}{2x-7}\)

Được xác định khi:

\(\left\{{}\begin{matrix}-x^2+5x-4\ge0\\2x-7\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\left(x-4\right)\left(x-1\right)\ge0\\2x\ne7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\left(x-4\right)\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}-\left(x-4\right)< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}-x\ge-4\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}-x< -4\\x< 1\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\le4\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le4\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

Điều kiện

\(3x-1\ge0\Leftrightarrow x\ge\dfrac{1}{3}\) 

\(4-x\ge0\Leftrightarrow x\le4\)

Kết hợp 2 đk \(\Rightarrow\dfrac{1}{3}\le x\le4\)

Bình phương 2 vế PT

\(4\left(3x-1\right)=4-x\)

\(\Leftrightarrow12x-4=4-x\Leftrightarrow13x=8\)

\(\Leftrightarrow x=\dfrac{8}{13}\) Đối chiếu với đk thỏa mãn

y = mx + n - 1 trùng y = (4 - n)x + 3 - n

\(\Rightarrow\left\{{}\begin{matrix}m=4-n\\n-1=3-n\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=4-m\\2n=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=2\\n=2\end{matrix}\right.\)

Vậy để 2 đường thẳng trùng nhau thì m = n = 2