\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)

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

\(=\sqrt{2^2}.\left|a-4\right|\)

\(=2\left(a-4\right)\)

\(=2a-8\)

\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)

\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)

\(=\sqrt{3^2}.\left|b-5\right|\)

\(=3\left(-b+5\right)\)

\(=-3b+15\)

Ta có: 

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\)

\(=\sqrt{\left(3x^2+6x+3\right)+9}+\sqrt{\left(5x^4-10x^2+5\right)+4}\)

\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\left(1\right)\)

Ta lại có:

\(-2x^2-4x+3=-2\left(x+1\right)^2+5\le5\left(2\right)\)

Từ (1) và (2) dấu = xảy ra khi \(x=-1\)

Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)

em ko bieets hu hu

#)Giải :

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

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

\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)

\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)

#)Giải :

b) Để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với đkxđ => 0 < x < 9

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

Bài 1:

Vì $a\geq 1$ nên:

\(a+\sqrt{a^2-2a+5}+\sqrt{a-1}=a+\sqrt{(a-1)^2+4}+\sqrt{a-1}\)

\(\geq 1+\sqrt{4}+0=3\)

Ta có đpcm

Dấu "=" xảy ra khi $a=1$

Bài 2:
ĐKXĐ: x\geq -3$

Xét hàm:

\(f(x)=x(x^2-3x+3)+\sqrt{x+3}-3\)

\(f'(x)=3x^2-6x+3+\frac{1}{2\sqrt{x+3}}=3(x-1)^2+\frac{1}{2\sqrt{x+3}}>0, \forall x\geq -3\)

Do đó $f(x)$ đồng biến trên TXĐ

\(\Rightarrow f(x)=0\) có nghiệm duy nhất

Dễ thấy pt có nghiệm $x=1$ nên đây chính là nghiệm duy nhất.

Bạn ơi bạn đã giải được bài 1 chưa vậy? 

\(Q=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)

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

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

\(=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}\)

Nếu  \(x\ge2\) thì 

\(Q=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x\sqrt{x-1}}{\left(x-2\right)\left(x-1\right)}=\frac{2x}{\left(x-2\right)\left(\sqrt{x-1}\right)}\)

Nếu \(x< 2\) thì \(Q=\frac{1-\sqrt{x-1}+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x}{\left(x-2\right)\left(x-1\right)}\)

Cảm ơn bạn nhiều nhưng mình thấy \(1-\frac{1}{x-1}=\frac{x-2}{x-1}\)  mà bạn sao lại bằng \(\frac{x}{x-1}\)được