3:
Chiều cao của tòa tháp là:
\(7\cdot tan61\simeq12,63\left(m\right)\)
2:
2:
a: \(\Leftrightarrow\sqrt{x+2}\cdot\dfrac{1}{3}+\sqrt{x+2}\cdot4-\dfrac{5}{2}\cdot\dfrac{3}{5}\cdot\sqrt{x+2}=\dfrac{3}{2}\)
=>\(\sqrt{x+2}\left(\dfrac{1}{3}+4-\dfrac{3}{2}\right)=\dfrac{3}{2}\)
=>\(\sqrt{x+2}=\dfrac{3}{2}:\dfrac{17}{6}=\dfrac{3}{2}\cdot\dfrac{6}{17}=\dfrac{9}{17}\)
=>x+2=81/289
=>x=81/289-2=-497/289
b: \(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3-8\sqrt{2x-3}+16}=5\)
=>\(\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}-4\right|=5\)
=>\(\sqrt{2x-3}+1+\left|\sqrt{2x-3}-4\right|=5\)
=>\(\sqrt{2x-3}+\left|\sqrt{2x-3}-4\right|=4\)(1)
TH1: x>=19/2
Phương trình (1) sẽ trở thành: \(\sqrt{2x-3}+\sqrt{2x-3}-4=4\)
=>\(2\sqrt{2x-3}=8\)
=>\(\sqrt{2x-3}=4\)
=>2x-3=16
=>2x=19
=>x=19/2(nhận)
TH2: 3/2<=x<19/2
Pt(1) sẽ là \(\sqrt{2x-3}+4-\sqrt{2x-3}=4\)
=>4=4(luôn đúng)
