Bài 1: Căn bậc hai

\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)

\(=\sqrt{6}+\sqrt{6}-1\)

\(=2\sqrt{6}-1\)

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\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)

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

\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)-3\left(\sqrt{2}-\sqrt{3}\right)}{-\sqrt{6}}\)

\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}\)

\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)

2:

a: =căn 17-4-căn 17=-4

b: =5-2căn 3-2căn 3=5-4căn 3

1:

a: =>|x+1|=-x

=>x<=0 và (x+1)^2=x^2

=>x<=0 và (x+1+x)(x+1-x)=0

=>x=-1/2

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

\(\Leftrightarrow\sqrt{\left(2x\right)^2-2\cdot2x\cdot1+1}=3\)

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

\(=\left|2x-1\right|=3\)

TH1: \(\left|2x-1\right|=2x-1\) với \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

Pt trở trành:

\(2x-1=3\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\left(tm\right)\)

TH2: \(\left|2x-1\right|=-\left(2x-1\right)\) với \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)

Pt trở thành:

\(-\left(2x-1\right)=3\) (ĐK: \(x< \dfrac{1}{2}\))

\(\Leftrightarrow-2x=2\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy: \(S=\left\{2;-1\right\}\)

Đk: \(x\in R\)

Pt \(\Leftrightarrow4x^2-4x+1=9\)

\(\Leftrightarrow4x^2-4x-8=0\)

\(\Leftrightarrow4x^2-8x+4x-8=0\)

\(\Leftrightarrow4x\left(x-2\right)+4\left(x-2\right)=0\)

\(\Leftrightarrow\left(4x+4\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{-1;2\right\}\)

ĐKXĐ: `x,y>0`

`P=(sqrtx+sqrty)/(sqrtx . sqrty)=1/(sqrtx)+1/(sqrty)`

Xét `f(a)=1/(sqrta)(a>0)` ta thấy `a` càng lớn `f(a)` càng nhỏ nên `P` không tồn tại `min`

1: =3-căn 2-3-căn 2=-2căn 2

2: \(=\sqrt{5}-2+3-\sqrt{5}=1\)

3: \(=\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)=7-4=3\)

4: \(=3+\sqrt{2}+2-\sqrt{2}=5\)

6: \(=\dfrac{2\sqrt{2}-\sqrt{6}}{2+\sqrt{3}+1}+\dfrac{2\sqrt{2}+\sqrt{6}}{2-\sqrt{3}+1}\)

\(=\dfrac{2\sqrt{2}-\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{6}}{3-\sqrt{3}}\)

\(=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+\sqrt{2}\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)

\(=\dfrac{\sqrt{2}\left(6-5\sqrt{3}+3\right)+\sqrt{2}\left(9+5\sqrt{3}\right)}{6}\)

\(=\dfrac{\sqrt{2}\cdot18}{6}=3\sqrt{2}\)

a: \(=\left(1+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{-\left(\sqrt{5}-1\right)}\right)\left(\sqrt{5}+1\right)\)

=(1-căn 5)(1+căn 5)

=1-5=-4

b: \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)

Lời giải:

a. $\sqrt{x^2}=1$

$\Leftrightarrow |x|=1$

$\Leftrightarrow x=\pm 1$

b. $\sqrt{4x^2-4x+1}=3$

$\Leftrightarrow \sqrt{(2x-1)^2}=3$
$\Leftrightarrow |2x-1|=3$

$\Leftrightarrow 2x-1=\pm 3$

$\Leftrightarrow x=-1$ hoặc $x=2$

3. ĐKXĐ: $x^2\geq 4$

$\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0$

Do $\sqrt{x^2-4}\geq 0; \sqrt{x^2+4x+4}\geq 0$ với mọi $x\in$ ĐKXĐ nên để tổng của chúng bằng $0$ thì:

$\sqrt{x^2-4}=\sqrt{x^2+4x+4}=0$

$\Leftrightarrow (x-2)(x+2)=(x+2)^2=0$

$\Leftrightarrow x=-2$

4. 

PT \(\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ x^2-4x+3=(x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x^2-4x+3=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x=6\end{matrix}\right.\Leftrightarrow x=3\)

Ý 1:

\(\sqrt{x^2}=1\\ \Leftrightarrow\left|x\right|=1\\ Vậy:x=1.hoặc.x=-1\\ S=\left\{\pm1\right\}\)

Ý 2:

\(\sqrt{4x^2-4x+1}=3\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\\ \Leftrightarrow\left|2x-1\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;2\right\}\)

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

=>căn x+2=0

=>x+2=0

=>x=-2

4: =>\(\left\{{}\begin{matrix}x>=3\\x^2-4x+3=x^2-6x+9\end{matrix}\right.\Leftrightarrow x=3\)

\(1,2\sqrt{27}+5\sqrt{12}-3\sqrt{48}\\ =2.3\sqrt{3}+5.2\sqrt{3}-3.4\sqrt{3}\\ =6\sqrt{3}+10\sqrt{3}-12\sqrt{3}\\ =4\sqrt{3}\)

\(2,\sqrt{147}+\sqrt{75}-4\sqrt{27}\\ =7\sqrt{3}+5\sqrt{3}-4.3\sqrt{3}\\ =7\sqrt{3}+5\sqrt{3}-12\sqrt{3}\\ =\sqrt{3}\left(7+5-12\right)\\ =0\)

\(3,3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\\ =3\sqrt{2}.\left(4-\sqrt{2}\right)+3\left(1-4\sqrt{2}+8\right)\\ =12\sqrt{2}-6+3-12\sqrt{2}+24\\ =21\)

\(4,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\\ =\sqrt{5}\left(2-5-4+11\right)\\ =4\sqrt{5}\)

1: =6căn 3+10căn 3-12căn 3=4căn 3

2: =7căn 3+5căn 3-12căn 3=0

3: =12căn 2-6+3(9-4căn 2)

=12căn 2-6+27-12căn 2=21

4: =2căn 5-5căn 5+4căn 5+9 căn 5

=10căn 5

=>x>=-2 và x^2+4=x^2+4x+4

=>4x=0 và x>=-2

=>x=0

a: =>x<=2x-1 và x>=0

=>-x<=-1 và x>=0

=>x>=1

b: =>x<=x+1 và x>=0

=>x>=0

c: =>x>=-4 và x^2-4x+6=x^2+8x+16

=>x>=-4 và -12x=10

=>x=-5/6(nhận)

d: =>căn x^2+1=x+3

=>x>=-3 và x^2+1=x^2+6x+9

=>6x=-8 và x>=-3

=>x=-4/3