b:

ĐKXĐ: x>0

 \(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

\(\Leftrightarrow x+1-2\sqrt{x}=0\)

=>x=1

a) ta có \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=-4x^2-4x+6\)

ta có : \(VT\ge\sqrt{16}+\sqrt{9}=7\) và \(VT\le7\)

\(\Rightarrow VT=VP\) \(\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

b) điều kiện \(x>0\)

ta có : \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

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

\(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\Leftrightarrow\dfrac{x+\sqrt{x}}{\sqrt{x}}=2\Leftrightarrow x+\sqrt{x}=2\sqrt{x}\)

\(\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=1\left(N\right)\end{matrix}\right.\)

vậy \(x=1\)

Mysterious Person giup mk nha

b:

ĐKXĐ: x>0

 \(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

\(\Leftrightarrow x+1-2\sqrt{x}=0\)

=>x=1

a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow x-3\le0\)

hay \(x\le3\)

b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)

\(\Leftrightarrow\left|2x-5\right|=5-2x\)

\(\Leftrightarrow2x-5\le0\)

hay \(x\le\dfrac{5}{2}\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2+4}+\sqrt{3\left(2x-1\right)^2+16}=6\)

Do \(\left(2x-1\right)^2\ge0\Rightarrow VT\ge\sqrt{0+4}+\sqrt{3.0+16}=6\)

Dấu "=" xảy ra khi và chỉ khi \(\left(2x-1\right)^2=0\)

\(\Rightarrow x=\frac{1}{2}\)

\(a,\sqrt{4x^2-20x+25}+2x=5\)

    \(\Rightarrow\sqrt{\left(2x-5\right)^2}+2x=5\)

  \(\Rightarrow4x=10\Rightarrow x=\frac{5}{2}\)

\(b,\sqrt{1-12x+36x^2}=5\)

  \(\Rightarrow6x-1=5\)

 \(\Rightarrow6x=6\Rightarrow x=1\) 

\(c,\sqrt{x^2+x}=x\)

  \(\Rightarrow x^2+x=x^2\)

\(\Rightarrow x=0\)   

\(c,\Rightarrow\left(x-2\right)^2-1=\left(x-2\right)^2\)

\(\Rightarrow-1=0\) (vô lý)

=> PT vô nghiệm