giúp voii mình cần gấpp

dấu nào là sao z b oi

AB =6cm

a, \(x-\sqrt{x-4\sqrt{x}+4}=8\)

\(\Leftrightarrow x-\sqrt{\left(\sqrt{x}-2\right)^2}=8\Leftrightarrow\left|\sqrt{x}-2\right|=x-8\)

ĐK : x >= 8

TH1   \(\sqrt{x}-2=x-8\Leftrightarrow x-\sqrt{x}-6=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2>0\right)=0\Leftrightarrow x=9\)

TH2 : \(\sqrt{x}-2=8-x\Leftrightarrow x+\sqrt{x}-10=0\)

\(\Leftrightarrow x_1==\frac{-1+\sqrt{41}}{2}\left(ktm\right);x_2=\frac{-1-\sqrt{41}}{2}\left(ktm\right)\)

Vậy x = 9 

\(x-\sqrt{x-4\sqrt{x}+4}=8\Leftrightarrow x-\sqrt{\left(\sqrt{x}-2\right)^2}=8\Leftrightarrow x-\left|\sqrt{x}-2\right|=8\)

Với 0 ≤ x < 4 pt <=> \(x+\sqrt{x}-2=8\Leftrightarrow x+\sqrt{x}-10=0\)(1)

Đặt t = √x ( t ≥ 0 ) (1) trở thành t² + t - 10 = 0 có Δ = 1 + 40 = 41 > 0 nên có hai nghiệm \(\hept{\begin{cases}t_1=\frac{-1+\sqrt{41}}{2}\left(tm\right)\\t_2=\frac{-1-\sqrt{41}}{2}\left(ktm\right)\end{cases}}\)

=> \(\sqrt{x}=\frac{-1+\sqrt{41}}{2}\Leftrightarrow x=\frac{21-\sqrt{41}}{2}\left(ktm\right)\)

Với x > 4 pt <=> \(x-\sqrt{x}+2=8\Leftrightarrow x-\sqrt{x}-6=0\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=0\\\sqrt{x}-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-2\left(voli\right)\\\sqrt{x}=3\end{cases}}\Leftrightarrow x=9\left(tm\right)\)

Vậy x = 9