∆ABC vuông tại A

⇒ tanC = AB : AC = 2 : 2,5 = 0,8

⇒ C ≈ 39⁰

⇒ ACD = 20⁰ + 39⁰ = 59⁰

∆ACD vuông tại A

⇒ tanACD = AD : AC

⇒ AD = AC.tanACD

= 2,5.tan59⁰

≈ 4,2 (m)

Độ dài vùng được chiếu sáng trên mặt đất:

BD = AD - AB = 4,2 - 2 = 2,2 (m)

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

Có phải đề bài như này không em?

Xem lại đề x + 3 hay x + 2 

\(\dfrac{1}{x^2\left(y-z\right)}=-\dfrac{3}{5}\Rightarrow x^2=-\dfrac{5}{3\left(y-z\right)}\)

\(\dfrac{1}{y^2\left(z-x\right)}=\dfrac{1}{3}\Rightarrow y^2=\dfrac{3}{\left(z-x\right)}\)

\(\dfrac{1}{z^2\left(x-y\right)}=3\Rightarrow z^2=\dfrac{1}{3\left(x-y\right)}\)

\(A=x^2.y^2.z^2=-\dfrac{5}{3\left(y-z\right)}.\dfrac{3}{z-x}.\dfrac{1}{3\left(x-y\right)}=\)

\(=-\dfrac{5}{3}.\dfrac{1}{\left(y-z\right)\left(z-x\right)\left(x-y\right)}=\)

a.

\(\sqrt{x^2-4x+1}=x\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-4x+1=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{4}\)

b.

\(\sqrt{5x^2-2x+2}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\5x^2-2x+2=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2-4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2}\)

c.

\(\sqrt{x^2-8x+16}=4-x\)

\(\Leftrightarrow\sqrt{\left(4-x\right)^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

\(\Leftrightarrow4-x\ge0\)

\(\Rightarrow x\le4\)

d.

\(\sqrt{3x+1}=\sqrt{4x-3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-3\ge0\\3x+1=4x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\end{matrix}\right.\)

\(\Rightarrow x=4\)

\(3\left(x^2+2x-1\right)-2\left(x^2+3x-1\right)+5x^2=0\)

=>\(3x^2+6x-3-2x^2-6x+2+5x^2=0\)

=>\(6x^2-1=0\)

=>\(6x^2=1\)

=>\(x^2=\dfrac{1}{6}\)

=>\(x=\pm\dfrac{\sqrt{6}}{6}\)