ĐKXĐ : \(x\ge0;y\ge1\)

\(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-1-6\sqrt{y-1}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=10\end{cases}}}\)

ĐK : \(x>1\) hoặc \(x\le-1\)

Ta có : \(\sqrt{\dfrac{x+1}{x-1}}=2\)

\(\Leftrightarrow\dfrac{x+1}{x-1}=4\)

\(\Leftrightarrow x+1=4x-4\)

\(\Leftrightarrow-3x=-5\)

\(\Leftrightarrow x=\dfrac{5}{3}\) ( Thỏa mãn )

Vậy \(x=\dfrac{5}{3}\)

Chúc bạn học tốt ...

ĐKXĐ x - 1 >0 <=> x>1
<=> \(\dfrac{x+1}{x-1}\) = 4 (bình phương cả 2 vế)
<=> x + 1 = 4x - 4
<=> x - 4x = - 4 -1
<=> -3x = -5
<=> x = 5/3 (TM)

ĐKXĐ: \(x\ge0\)

Đặt \(\sqrt{x}=a\)

\(\Rightarrow a^2-2a-1=0\)

\(\Rightarrow\left(a-1\right)^2=2\)

\(\Rightarrow\orbr{\begin{cases}a-1=\sqrt{2}\\a-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}a=\sqrt{2}+1\\a=-\sqrt{2}+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1< 0\left(v\text{ô}l\text{ý}\right)\end{cases}}}\Leftrightarrow x=\left(\sqrt{2}+1\right)^2=3+2.\sqrt{2}\)Vậy \(x=3+2.\sqrt{2}\)

P/S: Không chắc lắm

ĐK : \(x\ge2,y\ge3,z\ge4\) .

\(pt\Leftrightarrow x+y+z-6=2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-4}\)

\(\Leftrightarrow\left[\left(x-2\right)-2\sqrt{x-2}+1\right]+\left[\left(y-3\right)-2\sqrt{y-3}+1\right]+\left[\left(z-4\right)-2\sqrt{z-4}+1\right]=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-4}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\left(TM\right)\)

x = 0 ???

Đây là toán lớp 4?!

Bạn Nguyễn Đoan Hạnh cho mình bổ sung nhé 

Ư(9)={+-1;+-3;+-9}

Nếu x+1=-1 => x=-2

Nếu x+1=-3 => x = -4

Nếu X+1=-9 => x = -10

x+10 la boi cua x+1 

suy ra (x+1)+9 la boi cua x+1

suy ra 9 la boi cua x+1

U(9)={1;3;9}

Neu x+1=1 thi x=0

Neu x+1=3 thi x=2

Neu x+1=9 thi x=8

Vay x thuoc {0;2;8}