Kết quả là 6 nhé bạn

= 6 nha ban 

\(\sqrt{17+3\sqrt{32}}=\sqrt{17+12\sqrt{2}}=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)\(=\sqrt{\left(3+2\sqrt{2}\right)^2}=3+2\sqrt{2}\)

Biểu thức trở thành \(\sqrt{17-3\sqrt{32}+3+2\sqrt{2}}=\sqrt{20-10\sqrt{2}}\)

Bạn xem lại đề rồi đăng lại

= -2,112.760005

mình nhầm ạ

tách 2 căn rồi ạ

Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)

\(\Leftrightarrow A^3=4-3A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)

\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.

** Bài toán rút gọn**

Lời giải:

\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)

\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)

\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)

\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)

\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)

--------------------

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)

\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)

\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)

----------------------

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)

\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)

\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)

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

\(=-1\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(=6\)

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}\)

c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=2\sqrt{2}\)

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`

`=2/sqrt2=sqrt2`

`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`

`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`

`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`

`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`

`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`

`=(-2sqrt3)/sqrt2=-sqrt6`

`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`

`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`

`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`

`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`

`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`

`=(2sqrt3)/sqrt2=sqrt6`

`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`

`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`

`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`

`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`

`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`

`=2/sqrt2=sqrt2`

a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)

b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)

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

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

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

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

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

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2