ĐKXĐ:9x²-6x+1≥0

⇔(3x-1)²≥0

Mà (3x-1)² luôn luôn ≥0 với mọi x

⇒Để căn thức trên có nghĩa thì x∈R

1. ĐK \(x^2-8x+18\ge0\Rightarrow x^2-8x+16+2\ge0\)\(\Rightarrow\left(x-4\right)^2+2\ge2\forall x\)TXD : R
2. ĐK \(9x^2-6x+1>0\Rightarrow\left(3x-1\right)^2>0\forall x\ne\frac{1}{3}\)\(\Rightarrow TXD=R|\left\{\frac{1}{3}\right\}\)

Bài 1:

a: ĐKXĐ: \(\frac{x^2}{2x-1}\ge0\)

=>2x-1>0

=>2x>1

=>\(x>\frac12\)

b: \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}\cdot\sqrt[3]{\frac{1}{27}}\)

\(=\sqrt[3]{\frac{625}{5}}-\left(-6\right)\cdot\frac13=5+2=7\)

Bài 2:

a: \(\sqrt{\left(x+1\right)^2}=3\)

=>|x+1|=3

=>\(\left[\begin{array}{l}x+1=3\\ x+1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-4\end{array}\right.\)

b: ĐKXĐ: x>=-1

\(3\sqrt{4x+4}-\sqrt{9x+9}-8\cdot\sqrt{\frac{x+1}{16}}=5\)

=>\(3\cdot2\cdot\sqrt{x+1}-3\sqrt{x+1}-8\cdot\frac{\sqrt{x+1}}{4}=5\)

=>\(3\sqrt{x+1}-2\sqrt{x+1}=5\)

=>\(\sqrt{x+1}=5\)

=>x+1=25

=>x=24(nhận)

\(ĐK:x^2+1\ge0\Leftrightarrow x\in R\)

\(1+x^2\ge0\)(luôn đúng)

\(\Rightarrow x\in R\)

Để căn thức có nghĩa khi \(9x^2-6x+1\ge0\Leftrightarrow\left(3x-1\right)^2\ge0\Leftrightarrow x\ge\frac{1}{3}\)

DKXD \(\hept{\begin{cases}9x^2-6x+2\ge0\\x^2-5x-1\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(3x-1\right)^2+1\ge0\left(ld\right)\\x^2-5x-1\ge0\end{cases}\Leftrightarrow}x^2-5x-1\ge0\Leftrightarrow\orbr{\begin{cases}x\le\frac{5-\sqrt{29}}{2}\\x\ge\frac{5+\sqrt{29}}{2}\end{cases}}}\)