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

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

đặt \(\sqrt{x^2-2x+5}=a\left(a\ge0\right)\)

pt trở thành : \(a=a^2-6\)

\(\Leftrightarrow a^2-a-6=0\)

\(\Leftrightarrow\left(a-3\right)\left(a+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-2\left(loai\right)\end{cases}}\)

với \(a=3\Rightarrow\sqrt{x^2-2x+5}=3\Leftrightarrow x^2-2x+5=9\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4\cdot\left(-4\right)=4+16=20\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2+\sqrt{20}}{2}=\frac{2+2\sqrt{5}}{2}=1+\sqrt{5}\\x=\frac{2-\sqrt{20}}{2}=\frac{2-2\sqrt{5}}{2}=1-\sqrt{5}\end{cases}}\)

Ta có: \(x^2-2x+5=\left(x-1\right)^2+4\ge4\forall x\in R\)

Đặt \(t=\sqrt{x^2-2x+5}\left(t\ge2\right)\)

\(pt\Leftrightarrow t=t^2-6\Leftrightarrow t^2-t-6=0\Leftrightarrow t=-2\) (loại) hoặc t=3

Với t=3 \(\Leftrightarrow\sqrt{x^2-2x+5}=3\Leftrightarrow x^2-2x-4=0\Leftrightarrow x=1\pm\sqrt{5}\)