+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)    \(\left(ĐK:x\ge0\right)\)

        \(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)

        \(\Leftrightarrow3\sqrt{3x}=6\)

        \(\Leftrightarrow\sqrt{3x}=2\)

        \(\Leftrightarrow3x=4\)

        \(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{4}{3}\right\}\)

+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\)    \(\left(ĐK:x\ge1\right)\)

        \(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

        \(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)

        \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)

 Vậy \(S=\left\{1,15\right\}\)

+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\)       \(\left(ĐK:x\ge0\right)\)

         \(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

         \(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)

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

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

   Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)

    \(\Rightarrow\)\(\sqrt{x}-4< 0\)

   \(\Leftrightarrow\)\(\sqrt{x}< 4\)

   \(\Leftrightarrow\)\(x< 16\)

   Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)

 Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)

\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)  (Đk: x \(\ge\)0)

<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)

<=> \(3\sqrt{3x}=6\)

<=> \(\sqrt{3x}=2\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)

<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\) 

<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)

\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)

<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

<=>  \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)

<=> \(\sqrt{x}< 4\) <=> \(x< 16\)

Kết hợp với đk => S = {x|0 < x < 16}

khong biet

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\) 

sai đề

Phải là : y/(z+y+1)

xem lại đề: \(\frac{x}{x+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)

tuy nhiên đề thế nào làm vậy

hiển nhiên x=y=z=0 là nghiệm

2 số hạng đầu => x=y

\(\frac{x}{x+z+1}=\frac{z}{x+z-2}=\frac{2x+z}{3\left(x+z\right)}=2x+z\)

=> 2x+z=0=> x=z=0 (loại đang xét x, z khác 0)

xét 2x+z khác 0

<=> 3(x+z)=1=> x+z=1/3

\(2x+z=\frac{x+z}{2\left(x+z\right)-1}=\frac{\frac{1}{3}}{\frac{2.1}{3}-1}=-\frac{1}{3}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\z=1\end{cases}}\) Kết luận: \(\orbr{\begin{cases}x=y=z=0\\x=y=-\frac{2}{3};và,,,z=1\end{cases}}\)

a, (x-15):5+22=24

    ( x - 15 ) : 5 = 2

      x-15  = 10

       x = 25