\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}\)

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

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

#H

Đk:  x \(\ge\)-1

Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

<=> \(4\sqrt{x+1}-3\sqrt{x+1}=4\)

<=> \(\sqrt{x+1}=4\)

<=> x + 1 = 16

<=> x = 15

đk: \(x\ge-1\)

Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=4\)

\(\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x+1=4^2\Leftrightarrow x=15\)

a, D xác định

\(\Leftrightarrow\hept{\begin{cases}x\ge0;\sqrt{x}\ne0\\9-x\ne0\\x-3\sqrt{x}\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>0\\x\ne9\\x\ne0,x\ne9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)

Vậy ...

b, \(D=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(D=\left[\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(D=\left[\frac{3\sqrt{x}-x+x+9}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}\right]:\left[\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(D=\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(D=\frac{3}{3-\sqrt{x}}\cdot\left(-\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\right)\)

\(D=-\frac{3\sqrt{x}}{2\sqrt{x}+4}\)

Vậy ...

c, \(D< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}< -1\left(x>0;x\ne9\right)\)

\(\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+\frac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

\(\Leftrightarrow\frac{-\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

\(\Leftrightarrow-\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}>4\)

\(\Leftrightarrow x>16\left(tm\right)\)

Vậy ...

hàm số đồng biến khi \(a>0\)

\(< =>m>0\)thì hàm số dồng biến

ĐKXĐ: 2x-1 \(\ge\)0 \(\Leftrightarrow\)x\(\ge\)\(\frac{1}{2}\) 

  \(\frac{1}{3-x}\)\(\ge\)0 \(\Leftrightarrow\)3-x\(\le\)0\(\Leftrightarrow\)-x\(\le\)-3 \(\Leftrightarrow\)x\(\ge\)3

đK: \(\hept{\begin{cases}x\ge0\\x\ne9\\x\ne4\end{cases}}\)

đặt biểu thức là A

Ta có: \(A=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2-\sqrt{x}\right)+9-x}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)

\(=\left[\frac{3}{\sqrt{x}+3}\right]:\left[\frac{4\sqrt{x}-x-4}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)

\(=\left[\frac{3}{\sqrt{x}+3}\right].\left[\frac{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{-\left(2-\sqrt{x}\right)^2}\right]=\frac{3}{\sqrt{x}-2}\)

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