Ôn tập chương 1: Căn bậc hai. Căn bậc ba

1) \(\sqrt{-8x}\)

Xác định khi:

\(-8x\ge0\)

\(\Leftrightarrow x\le0\)

2) \(\sqrt{\dfrac{1}{4}}-2a=\dfrac{1}{2}-2a\)

Được xác đinh với mọi a

3) \(\sqrt{x^2-9}\)

Xác định khi:

\(x^2-9\ge0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\x-3\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+3\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\x\le3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

4) \(\sqrt{\dfrac{-3}{2+x}}\) 

Xác định khi:

\(\dfrac{-3}{2+x}\ge0\) và \(2+x\ne0\) Mà: \(-3< 0\)

\(\Leftrightarrow2+x< 0\)

\(\Leftrightarrow x< -2\)

5) \(\sqrt{\dfrac{\sqrt{6}-4}{m+2}}\)

Xác định khi: 

\(\dfrac{\sqrt{6}-4}{m+2}\ge0\) và \(m+2\ne0\)

Mà: \(\sqrt{6}-4< 0\)

\(\Rightarrow m+2< 0\)

\(\Leftrightarrow m< -2\)

1: ĐKXĐ: -8x>=0

=>x<=0

2: Sửa đề: \(\sqrt{\dfrac{1}{4}-2a}\)

ĐKXĐ: 1/4-2a>=0

=>2a<=1/4

=>a<=1/8
3: ĐKXĐ: x^2-9>=0

=>(x-3)(x+3)>=0

=>x>=3 hoặc x<=-3

4: ĐKXĐ: -3/x+2>=0

=>x+2<0

=>x<-2

5: ĐKXĐ: \(\dfrac{\sqrt{6}-4}{m+2}>=0\)

=>m+2<0

=>m<-2

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot\sqrt{2}+b=4-\sqrt{2}\\2a+b=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\left(\sqrt{2}-2\right)=4-2\sqrt{2}\\2a+b=\sqrt{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{4-2\sqrt{2}}{\sqrt{2}-2}=-2\\b=\sqrt{2}+4\end{matrix}\right.\)

ĐKXĐ: x>=0

1: P=2

=>\(2\left(\sqrt{x}+3\right)=\sqrt{x}+2\)

=>\(2\sqrt{x}+6=\sqrt{x}+2\)

=>\(\sqrt{x}=-4\)(vô lý)

2: P>2

=>P-2>0

=>\(\dfrac{\sqrt{x}+2-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}>0\)

=>\(\sqrt{x}+2-2\left(\sqrt{x}+3\right)>0\)

=>\(-\sqrt{x}-4>0\)

=>\(\sqrt{x}+4< 0\)(vô lý)

4:

a: \(P=\left(\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x}\)

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

b: Khi x=3-2căn 2 thì \(P=-\left(\sqrt{2}-1\right)+1=-\sqrt{2}+1+1=2-\sqrt{2}\)

c: \(P\cdot\sqrt{x}=-x+\sqrt{x}\)

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

Dấu = xảy ra khi x=1/4

giải gúp em bài 2 vs ạ

\(\dfrac{\sqrt{y}}{\sqrt{xy}-x}-\dfrac{\sqrt{x}}{y-\sqrt{xy}}\)

\(=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{y}-\sqrt{x}\right)}=\dfrac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}}\)

b: Khi x=16 thì \(A=\dfrac{3\cdot4-9}{4+3}=\dfrac{3}{7}\)

c: A<-1

=>A+1<0

=>\(\dfrac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\)

=>\(4\sqrt{x}-6< 0\)

=>\(\sqrt{x}< \dfrac{3}{2}\)

=>0<=x<9/4

a: \(R=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

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

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

b: Khi x=4+2căn 3 thì \(R=\dfrac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1-2\right)}=\dfrac{3+\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{3+\sqrt{3}}{2}\)

c: Để R>0 thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)

=>\(\sqrt{x}-2>0\)

=>x>4

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

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

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

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

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)