a,Để \(\sqrt{x^2-8x-9}\) có nghĩ thì

 \(x^2-8x-9\ge0\)

\(\Leftrightarrow x^2+x-9x-9\ge0\)

\(\Leftrightarrow x\left(x+1\right)-9\left(x+1\right)\ge0\)

\(\Leftrightarrow\left(x+1\right)\left(x-9\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-1\\x\ge9\end{cases}\Rightarrow}x\ge9\)

\(or\orbr{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\le-1\\x\le9\end{cases}\Rightarrow}x\le-1\)

\(Để\sqrt{4-9x^2}\text{có nghĩa}\)

\(\Rightarrow4-9x^2\ge0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)\ge0\)

\(\Leftrightarrow-\frac{2}{3}\le x\le\frac{2}{3}\)

Bài 1:

Căn bậc hai số học của \(\left(-7\right)^2\) là |-7|=7

Bài 2:

a: \(0,2\cdot\sqrt{\left.\left(-10\right)^2\right.\cdot3}+2\cdot\sqrt{\left(\sqrt5-\sqrt3\right)^2}\)

\(=0,2\cdot10\cdot\sqrt3+2\cdot\left(\sqrt5-\sqrt3\right)\)

\(=2\sqrt3+2\sqrt5-2\sqrt3=2\sqrt5\)

Bài 3:

\(\sqrt{\left(2x-1\right)^2}-5=0\)

=>\(\left|2x-1\right|-5=0\)

=>|2x-1|=5

=>\(\left[\begin{array}{l}2x-1=5\\ 2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=6\\ 2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

Câu 4:

ĐKXĐ: 4-3x>=0

=>3x<=4

=>\(x\le\frac43\)

Để căn thức \(\sqrt{\dfrac{2x+1}{x^2+1}}\) có nghĩa thì:

\(\left\{{}\begin{matrix}\dfrac{2x+1}{x^2+1}\ge0\\x^2+1\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1\ge0\left(vì.x^2+1>0\forall x\right)\\x^2+1\ne0\forall x\end{matrix}\right.\)

\(\Rightarrow2x\ge-1\Leftrightarrow x\ge-\dfrac{1}{2}\)

#\(Toru\)

\(\sqrt{\dfrac{2x+1}{x^2+1}}\)

Có nghĩa khi:

\(\dfrac{2x+1}{x^2+1}\ge0\)

\(\Leftrightarrow2x+1\ge0\)

\(\Leftrightarrow2x\ge-1\)

\(\Leftrightarrow x\ge-\dfrac{1}{2}\)

Vậy: ... 

a: ĐKXĐ: x-2>=0 và 6-2x>=0

=>2<=x<=3

b: DKXĐ: x+2>=0

=>x>=-2