a) \(\sqrt{1-x}+\sqrt{x+4}=3\left(Đk:-4\le x\le1\right)\)

\(1-x+x+4+2\sqrt{\left(1-x\right)\left(x+4\right)}=9\)

\(\sqrt{-x^2-3x+4}=2\)

\(-x^2-3x+4=4\)

\(x^2+3x=0\)

\(x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\left(TM\right)\\x=0\left(TM\right)\end{matrix}\right.\)

b) \(x^2+4x+5=2\sqrt{2x+3}\) \(\left(Đk:x\ge\dfrac{-3}{2}\right)\)

\(\)\(x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\)

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

Vì \(\left(x+1\right)^2;\left(\sqrt{2x+3}-1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+3=1\end{matrix}\right.\)\(\Leftrightarrow x=-1\)(TM)

\(P=\dfrac{\sqrt{a}}{a\sqrt{a}-3a+3\sqrt{a}+1}\) \(\left(a\ge0\right)\)

   \(=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)^3-1}\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\left(\sqrt{a}-1\right)^3-1\ge-2\end{matrix}\right.\)

\(\Rightarrow P\le0\)

\(MaxP=0\Leftrightarrow a=0\)

\(\widehat{ABC}=90^o+15^o30'=105,5^o\)

\(\widehat{CAB}=90^o-45^o=45^o\)

\(\widehat{ACB}=180^o-105,5^o-45^o=29,5^o\)

Ta có:

\(\dfrac{AC}{sin105,5}=\dfrac{AB}{sin29,5}\)

\(\Rightarrow AC=\dfrac{120.sin105,5}{sin29,5}=234,8\left(m\right)\)

Ta có:

\(sin45=\dfrac{CH}{AC}\)

\(\Rightarrow CH=234,8.sin45=166\left(m\right)\)

\(AB^2=AC^2+BC^2-2AC.BC.cosACB\)

          \(=700^2+1000^2-2.700.1000.cos120\)

          \(=2190000\)

\(\Rightarrow AB\simeq1480\left(m\right)\)

\(AB=\sqrt{AH^2+HB^2}=5\sqrt{10}m\)

\(cos\widehat{ABH}=\dfrac{AB^2+HB^2-AH^2}{2AB.HB}=0,95\)

\(\Rightarrow\widehat{ABH}=18,2^o\)

\(\Rightarrow\widehat{ABC}=180^o-45^o-\left(90^o-18,2^o\right)=63,2^o\)

\(\Rightarrow\widehat{ACB}=180^o-45^o-63,2^o=71,8^o\)

Ta có:

\(\dfrac{BC}{sin45}=\dfrac{AB}{sin71,8}\)

\(\Rightarrow BC=\dfrac{5\sqrt{10}.sin45}{sin71,8}\simeq11,77\left(m\right)\)

1h người đi xe đạp đi được: \(S_1=10.1=10km\)

1h30' người đi bộ đi được: \(S_2=\dfrac{5.3}{2}=7,5km\)

Hai người cách nhau:

\(10+7,5=17,5km\)

Gọi t là thời gian 2 người gặp nhau (t>0)

Ta có phương trình:

\(10t-5t=17,5\)

\(t=3,5\left(TM\right)\)

Vậy sau 5h người đi xe đạp đuổi kịp người đi bộ

Ta có:

\(\widehat{MNB}=180^o-45^o=135^o\)

\(\Rightarrow\widehat{MBN}=180^o-135^o-30^o=15^o\)

Ta có:

\(\dfrac{MN}{sin15}=\dfrac{NB}{sin30}\)

\(\Rightarrow NB=\dfrac{500.sin30}{sin15}\simeq966m\)

Ta có:

\(sin45=\dfrac{AB}{NB}\)

\(\Rightarrow AB=sin45.NB=sin45.966\simeq683,07m\)

Ta có:
\(S_{ABC}=\dfrac{1}{2}h_a.a\)

\(S_{ABC}=\dfrac{1}{2}b.c.sinA\)

\(\Rightarrow\left\{{}\begin{matrix}h_a=\dfrac{2S_{ABC}}{a}\\c.sinA=\dfrac{2S_{ABC}}{b}\end{matrix}\right.\)

Để \(h_a=c.sinA\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\) Tam giác ABC cân tại C