Thời gian để hai xe gặp nhau là: \(\dfrac{140}{30+40}=2\) (giờ)

Hai xe gặp nhau lúc 9h 30ph.

Vị trí gặp nhau cách A: \(40.2=80\left(km\right)\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{\sqrt{32}-\sqrt{12}}{-\left(\sqrt{48}-\sqrt{18}\right)}=-\dfrac{\sqrt{2}\left(\sqrt{16}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{16}-\sqrt{6}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}=\dfrac{-\sqrt{6}}{3}\)

ĐKXĐ: \(x\ge-3\).

\(PT\Leftrightarrow2\left(x^2+18\right)=7\sqrt{\left(x+3\right)\left(x^2-3x+9\right)}\). (*)

Đặt \(\sqrt{x+3}=a;\sqrt{x^2-3x+9}=b\left(a,b\ge0\right)\).

\(\left(\cdot\right)\Leftrightarrow2\left(b^2+3a^2\right)=7ab\Leftrightarrow6a^2-7ab+2b^2=0\)

\(\Leftrightarrow\left(3a-2b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}3a=2b\\2a=b\end{matrix}\right.\).

+) \(3a=2b\Leftrightarrow3\sqrt{x+3}=2\sqrt{x^2-3x+9}\Leftrightarrow4\left(x^2-3x+9\right)=9\left(x+3\right)\Leftrightarrow4x^2-12x+36=9x+27\Leftrightarrow4x^2-21x+9=0\Leftrightarrow x=\dfrac{21\pm3\sqrt{33}}{8}\). (TMĐK)

+) \(2a=b\Leftrightarrow4\left(x+3\right)=x^2-3x+9\Leftrightarrow x^2-7x-3=0\Leftrightarrow x=\dfrac{7\pm\sqrt{61}}{2}\left(TMĐK\right)\). 

Vậy...

\(PT\Leftrightarrow x\left(x^2-9\right)-\left(x^3+8\right)=0\Leftrightarrow-9x-8=0\Leftrightarrow x=\dfrac{-8}{9}\)

Gọi số cần tìm là \(\overline{ab}\left(a,b\in N;5\le a,b\le9\right)\).

Theo bài ra ta có a + b = 14.

\(\overline{ba}-\overline{ab}=18\Leftrightarrow\left(10b+a\right)-\left(10a+b\right)=18\Leftrightarrow9\left(b-a\right)=18\Leftrightarrow b-a=2\).

Kết hợp với a + b = 14 ta có \(b=\dfrac{14+2}{2}=8;a=6\). (TMĐK)

Vậy số cần tìm là 68.

Ta có \(\Delta=5^2-12\left(\sqrt{3}-3\right)=61-12\sqrt{3}\).

Do đó \(x=\dfrac{\pm\sqrt{61-12\sqrt{3}}-5}{6}\)

Ta có \(x^2-\left|x\right|\ge0\Leftrightarrow\left|x\right|^2\ge\left|x\right|\Leftrightarrow\left|x\right|\ge1\).

Khi đó \(PT\Leftrightarrow\sqrt{\left(\left|x\right|-1\right)^2}=x^2-\left|x\right|\)

\(\Leftrightarrow\left|x\right|-1=x^2-\left|x\right|\Leftrightarrow\left(\left|x\right|-1\right)^2=0\Leftrightarrow\left|x\right|=1\Leftrightarrow x=\pm1\) (thỏa mãn |x| \(\geq \) 1).

Vậy...

a) Xét \(\Delta ABD\) và \(\Delta AND\) có AD - cạnh chung, \(\widehat{ABD}=\widehat{AND}=90^o\), \(\widehat{BAD}=\widehat{NAD}\) (gt)

Do đó \(\Delta ABD=\Delta AND\left(ch-gn\right)\).

b) Ta có \(DI>DB=ND\)