Căn bậc hai - Hỏi đáp

So sánh:

1) \(2\sqrt{27}\) và \(\sqrt{147}\)

+ \(2\sqrt{27}\) = \(6\sqrt{3}\)

+ \(\sqrt{147}\) = \(7\sqrt{3}\)

⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)

Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)

2) \(2\sqrt{15}\) và \(\sqrt{59}\)

+ \(2\sqrt{15}\) = \(\sqrt{60}\)

⇒ \(\sqrt{60}\) > \(\sqrt{59}\)

Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)

3) \(2\sqrt{2}-1\) và 2

\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)

So sánh: 3 và \(2\sqrt{2}\)

+ 3 = \(\sqrt{9}\)

+ \(2\sqrt{2}=\sqrt{8}\)

⇒ \(\sqrt{8}\) < \(\sqrt{9}\)

⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1

⇒ \(2\sqrt{2}\) - 1 < 3 - 1

Vậy: \(2\sqrt{2}-1< 2\)

4) \(\frac{\sqrt{3}}{2}\) và 1

+ 1 = \(\frac{2}{2}\)

⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)

Vậy: \(\frac{\sqrt{3}}{2}\) < 1

5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)

+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)

⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)

Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)

a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)

=-7

b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)

1) ta có : \(2\sqrt{5}=\sqrt{20}\) và \(5\sqrt{2}=\sqrt{50}\)

mà \(\sqrt{20}< \sqrt{50}\) \(\Rightarrow\) \(2\sqrt{5}-5\sqrt{2}\) \(< 0\)

mà \(1>0\)

\(\Rightarrow\) \(2\sqrt{5}-5\sqrt{2}< 1\)

2) ta có : \(\sqrt{25}< \sqrt{26}\) \(\Rightarrow\) \(\sqrt{25}-\sqrt{26}\) \(< 0\)

mà \(\sqrt{9}>0\) \(\Rightarrow\) \(\sqrt{9}>\sqrt{25}-\sqrt{26}\)

1, Ta thấy \(2\sqrt{5}-5\sqrt{2}< 0\)

\(1>0\)

\(\Leftrightarrow2\sqrt{5}-5\sqrt{2}< 1\)

Vậy..

2, Ta thấy \(\sqrt{9}=3>0\)

\(\sqrt{25}-\sqrt{26}< 0\)

\(\Leftrightarrow\sqrt{9}>\sqrt{25}-\sqrt{26}\)

Vậy...

\(x-4\sqrt{x^2+3}=0\)

\(\Leftrightarrow x=4\sqrt{x^2+3}\)

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

\(\Leftrightarrow x^2=16\left(x^2+3\right)\)

\(\Leftrightarrow x^2=16x^2+48\)

\(\Leftrightarrow15x^2+48=0\)

\(\Leftrightarrow15x^2=-48\) (vô lí)

Vậy...

Ta có :

\(x< -3\Leftrightarrow x+3< 0\)

Lại có :

\(3x+\sqrt{9+6x+x^2}\)

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

\(=3x+\left|x+3\right|\)

\(=3x-x-3\) (Vì \(x+3< 0\) )

\(=2x-3\)

Vậy...

đăng từng bài một lên thui bạn ơi ;'<

a/ \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|+\left|2\sqrt{3}-\sqrt{2}\right|=\sqrt{3}+\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}\) (Vì \(\sqrt{3}+\sqrt{2}>0\) , \(2\sqrt{3}-\sqrt{2}>0\) )

b/ \(\sqrt{5-2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}-\sqrt{2}\right|+\left|2\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}-2\sqrt{2}\)

c/ \(11-\sqrt{33}=\sqrt{11}.\sqrt{11}-\sqrt{3}.\sqrt{11}=\sqrt{11}\left(\sqrt{11}-\sqrt{3}\right)\)

1/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Thay \(x=9\) vào biểu thức A ta có :

\(A=\frac{\sqrt{9}+1}{\sqrt{9}-1}=\frac{3+1}{3-1}=2\)

Vậy...

2/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có :

\(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)

Vậy....

b/ Ta có :

\(2P=2\sqrt{x}+5\)

\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\)

\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\)

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

\(\Leftrightarrow2\sqrt{x}-1=0\)

\(\Leftrightarrow x=\frac{1}{4}\)

Vậy..

Câu 1 :

\(A=\frac{\sqrt{x}+1}{\sqrt{x-1}}\) khi x = 9

tại x = 9 thay vào A ta được : \(\frac{\sqrt{9}+1}{\sqrt{9}-1}\) = \(\frac{3+1}{3-1}=\frac{4}{2}=2\)

Câu 2 :

a, Ta có : P = \(\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2}{\sqrt{x}.\sqrt{x}+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x+2}\right)}+\frac{1}{\sqrt{x}+2}\right)\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

=\(\frac{x-2+2\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}\) => đpcm

a/\(x+2\sqrt{x}=\sqrt{x}\left(\sqrt{x}+2\right)\)

b/\(3x-2\sqrt{x}=\sqrt{x}\left(3\sqrt{x}-2\right)\)

c/\(2\sqrt{x}-4x=2\sqrt{x}\left(1-2\sqrt{x}\right)\)

d/\(4x-3\sqrt{x}=\sqrt{x}\left(4\sqrt{x}-3\right)\)

Chúc bạn học tốt nhé

a, Ta có : \(x+2\sqrt{x}\)\(=\sqrt{x}\left(\sqrt{x}+2\right)\)

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

c, Ta có : \(2\sqrt{x}-4x=2\sqrt{x}\left(1-2\sqrt{x}\right)\)

d, Ta có : \(4x-3\sqrt{x}=\sqrt{x}\left(4\sqrt{x}-3\right)\)

ĐKXĐ : \(\left\{{}\begin{matrix}\frac{x-1}{x}\ge0\\x\ne0\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x< 0\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x\le1\\x< 0\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}x\ge1\\x< 0\end{matrix}\right.\)

Với \(x\ge1\)

=> \(x\sqrt{\frac{x-1}{x}}\ge0\)

=> \(x-2\ge0\)

=> \(x\ge2\)

=> \(\left[{}\begin{matrix}x\ge2\\x< 0\end{matrix}\right.\)

Ta có : \(x\sqrt{\frac{x-1}{x}}=x-2\)

=> \(x^2\left(\frac{x-1}{x}\right)=x^2-4x+4\)

=> \(x\left(x-1\right)=x^2-4x+4\)

=> \(x^2-4x+4=x^2-x\)

=> \(3x=4\)

=> \(x=\frac{4}{3}\left(L\right)\)

Vậy phương trình vô nghiệm .

a) \(\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}\)

\(1^2=1\)

vậy \(\left(\frac{\sqrt{3}}{2}\right)^2< 1^2=>\frac{\sqrt{3}}{2}< 1\)

b) \(\left(-\frac{\sqrt{10}}{2}\right)^2=\frac{5}{2}\)

\(\left(-2\sqrt{5}\right)^2=20\)

vậy \(\left(-\frac{\sqrt{10}}{2}\right)^2< \left(2\sqrt{5}\right)^2=>\frac{\sqrt{10}}{2}< 2\sqrt{5}=>-\frac{\sqrt{10}}{2}>-2\sqrt{5}\)

a) Ta có: \(\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{4}}\)

mà \(\sqrt{\frac{3}{4}}< \sqrt{\frac{4}{4}}=\sqrt{1}=1\)

nên \(\frac{\sqrt{3}}{2}< 1\)

b) Ta có: \(\frac{\sqrt{10}}{2}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}}=\sqrt{2.5}\)

\(2\sqrt{5}=\sqrt{4\cdot5}=\sqrt{20}\)

mà \(\sqrt{2.5}< \sqrt{20}\)(vì 2,5<20)

nên \(\frac{\sqrt{10}}{2}< 2\sqrt{5}\)

\(\Leftrightarrow-\frac{\sqrt{10}}{2}>-2\sqrt{5}\)

a, ĐKXĐ : \(2x-3\ge0\)

=> \(x\ge\frac{3}{2}\)

Ta có : \(P=x-2\sqrt{2x-3}\)

- Đặt \(t=\sqrt{2x-3}\left(t\ge0\right)\)

=> \(t^2=2x-3\)

=> \(x=\frac{t^2+3}{2}\)

- Thay vào P ta được : \(P=\frac{t^2+3}{2}-2t\)

b, Ta có : \(P=\frac{t^2+3-4t}{2}\)

=> \(P=\frac{t^2-4t+4-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)

Ta thấy : \(\left(t-2\right)^2\ge0\forall x\)

=> \(\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\ge-\frac{1}{2}\forall x\)

Vậy \(Min_P=-\frac{1}{2}\) <=> \(t-2=0\)

<=> \(t=2\left(TM\right)\)

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

<=> \(2x-3=4\)

<=> \(2x=7\)

<=> \(x=\frac{7}{2}\left(TM\right)\)

a) Xét \(\Delta=4-4.\left(-m\right)\)

Vì pt có nghiệm kép nên \(\Delta=0\Leftrightarrow m=-1\)

b) Xét \(\Delta=\left(-3\right)^2-4m=9-4m\)

Vì pt có nghiệm kép nên \(\Delta=0\Leftrightarrow m=\frac{9}{4}\)

a/ Xét phương trình :

\(x^2+2x-m=0\)

Ta có :

\(\Delta'=1-\left(-m\right)=m+1\)

Để phương trình có nghiệm kép :

\(\Leftrightarrow\Delta=0\Leftrightarrow m+1=0\Leftrightarrow m=-1\)

Vậy..

b/ Xét phương trình :

\(x^2-3x+m=0\)

Ta có :

\(\Delta=\left(-3\right)^2-4m=9-4m\)

Để phương trình có nghiệp kép

\(\Leftrightarrow\Delta=0\Leftrightarrow9-4m=0\Leftrightarrow m=\frac{9}{4}\)

Vậy...