Question

câu 1:Với giá trị nào của X thì căn thức sau có nghĩa

a)\(\sqrt{\dfrac{4}{2x+3}}\),b)\(\sqrt{_{-\dfrac{2}{x+1}}}\),c)\(\dfrac{x}{x+2}\)+\(\sqrt{x+2}\),d)\(\dfrac{x}{x2-4}\)+\(\sqrt{x-2}\),e)\(\sqrt{9x2-6x+1}\)

Answer 1

\(a,2x+3>0\\ 2x>-3\\ x>-\dfrac{3}{2}\\ b,x+1< 0\\ x< -1\\ c,x>-2\\ d,x>2\)

e, x ≠ 3

Answer 2

`a)`\(\sqrt{\dfrac{4}{2x+3}}=\dfrac{\sqrt{4}}{\sqrt{2x+3}}=\dfrac{2}{\sqrt{2x+3}}\)

\(ĐK:x>-\dfrac{3}{2}\)

`b)`\(\sqrt{-\dfrac{2}{x+1}}=\dfrac{\sqrt{2}}{\sqrt{-x-1}}\)

\(ĐK:x\le-1\)

`c)`\(\dfrac{x}{x+2}+\sqrt{x+2}\)

\(ĐK:x\ne-2\)

         \(x\ge-2\)

\(\Rightarrow x>-2\)

`d)`\(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\sqrt{x-2}\)

\(ĐK:x\ne\pm2\)

         \(x\ge2\)

\(\Rightarrow x>2\)

`e)`\(\sqrt{9x^2-6x+1}=\sqrt{\left(3x-1\right)^2}\)

\(ĐK:x\in R\)

Answer 3

a) \(\sqrt{\dfrac{4}{2x+3}}\)

`ĐKXĐ : 2x+3 > 0`

`<=> 2x > -3`

`<=> x > -3/2`

b)\(\sqrt{-\dfrac{2}{x+1}}\)

ĐKXĐ : `x+1 <0`

`<=> x < -1 `

c)\(\dfrac{x}{x+2}+\sqrt{x+2}\)

ĐKXĐ : ` x+2>0`

`<=> x>-2`

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

ĐK : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ge-2\end{matrix}\right.\)

e)\(\sqrt{9x^2-6x+1}=\sqrt{9x^2-3x-3x+1}\)

\(=\sqrt{3x\left(3x-1\right)-\left(3x-1\right)}=\sqrt{\left(3x-1\right)^2}\)

ĐKXĐ : `x∈R`