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

Với \(a>0;a\ne1\)

\(P=\left(\frac{1-a\sqrt{a}}{a-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{a+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left(\frac{-\left(\sqrt{a}-1\right)\left(1+\sqrt{a}+a\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}+\sqrt{a}\right)\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}-\sqrt{a}\right)\)

\(=\left(\frac{-a-\sqrt{a}-1+a}{\sqrt{a}}\right)\left(\frac{a-\sqrt{a}+1-a}{\sqrt{a}}\right)=\left(\frac{-\sqrt{a}-1}{\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{\sqrt{a}}\right)\)

\(=-\frac{1-a}{a}=\frac{a-1}{a}\)

à thêm a,b,c>0 nha 

Theo BĐT Svacxo có : \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

\(< =>1\ge\frac{\left(a+b+c\right)^2}{3}< =>\left(a+b+c\right)^2\le3< =>a+b+c\le\sqrt{3}\)

Dấu "=" xảy ra \(< =>a=b=c=\frac{1}{\sqrt{3}}\)

a, Với \(a>0;a\ne1\)

\(P=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\right)\)

\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{-4\sqrt{a}}{a-1}\right)=\frac{-4\sqrt{a}\left(a-1\right)^2}{4a\left(a-1\right)}=\frac{1-a}{\sqrt{a}}\)

b, Ta có : \(P< 0\Rightarrow\frac{1-a}{\sqrt{a}}< 0\Rightarrow1-a< 0\Leftrightarrow a>1\)

Ta có

  \(P=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)  Điều kiện \(a\ge0,a\ne\pm\sqrt{2}\)

      \(=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{1}{\sqrt{a}-2}\)

       \(=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-1\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

       \(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

       \(=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

       \(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)

Viết đề bài khó hiểu quá!

lớp 9 mà lị

này thì Cauchy cái gì bạn :v

Với x ≥ 0 thì √x + 5 ≥ 5 => 10/(√x + 5) ≤ 2 => -10/(√x + 5) ≥ -2

Dấu "=" xảy ra <=> x = 0 . Vậy MinA = -2