#)Giải :

a)      A = √(3+√5)-√(3-√5)-√2 

<=>A√2=√(6+2√5)-√(6-2√5)-2

<=>A√2=√(√5+1)^2-√(√5-1)-2

<=>A√2=√5+1-√5+1-2

<=>A√2=0

<=>A=0

=>√(3+√5)-√(3-√5)-√2 =0

b)       B=√(4-√7)-√ (4+√7)+√7

<=>B√2=√(8-2√7)-√(8+2√7)+2√7

<=>B√2=√(√7-1)^2-√(√7+1)^2+2√7

<=>B√2=√7-1-√7-1+2√7

<=>B√2=2√7-2

<=>B=(2√7-2)/√2

=√14-√2

                      #~Will~be~Pens~3

Câu a) hình như sai đề đúng không bạn ?

b) \(B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)

Xét \(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)

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

\(=8-2\sqrt{16-7}\)

\(=8-2\cdot3\)

\(=2\)

\(\Rightarrow\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=-\sqrt{2}\)( vì \(\sqrt{4-\sqrt{7}}< \sqrt{4+\sqrt{7}}\))

Khi đó : \(B=-\sqrt{2}+\sqrt{7}\)

Góp ý nhẹ với bạn ๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ ( Team TST 14 ) là không biết thì đừng làm nhé 

a)

\(\sqrt{3-\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}}{2}=\sqrt{\frac{5-2\sqrt{5}.1+1}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{5}-1\right)^2}{2}}=\frac{\sqrt{10}-\sqrt{2}}{2}\)

Tương tự ta cũng có: \(\sqrt{3+\sqrt{5}}=\frac{\sqrt{10}+\sqrt{2}}{2}\)

Suy ra \(A=\sqrt{\frac{\sqrt{10}+\sqrt{2}}{2}-\frac{\sqrt{10}-\sqrt{2}}{2}}-\sqrt{2}\)

\(=\sqrt[4]{2}-\sqrt{2}\)

Có gì đó sai sai ạ!

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

\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)

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

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

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

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

Tham khảo nha e

đăng câu hỏi kiểu j mà đặng đc lên như thế này đấy

1.

Đặt \(\sqrt[3]{2+\sqrt{b}}=x;\sqrt[3]{2-\sqrt{b}}=y\)

Do \(x>0\Rightarrow x^2+y^2-xy=\dfrac{3}{4}x^2+\left(\dfrac{1}{2}x-y\right)^2>0\)

\(PT\Leftrightarrow\dfrac{x^3+y^3}{a}+xy=x^2+y^2\Leftrightarrow\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{a}=x^2-xy+y^2\\ \Leftrightarrow\left(x^2-xy+y^2\right)\left(\dfrac{x+y}{a}-1\right)=0\\ \Leftrightarrow\dfrac{x+y}{a}=1\\ \Leftrightarrow\sqrt[3]{2+\sqrt{b}}+\sqrt[3]{2-\sqrt{b}}=a\left(1\right)\\ \Leftrightarrow\left(\sqrt[3]{2+\sqrt{b}}+\sqrt[3]{2-\sqrt{b}}\right)^3=a^3\\ \Leftrightarrow4+3a\sqrt[3]{4-b}=a^3\left(2\right)\\ \Rightarrow4-b=\left(\dfrac{a^3-4}{3a}\right)^3\)

Mặt khác \(b\in \mathbb{Z^+}\)

\(\Rightarrow\left(a^3-4\right)⋮3a\Rightarrow\left(a^3-4\right)⋮a\\ \Rightarrow4⋮a\Rightarrow a\in\left\{1;2;4\right\}\)

Với \(a=1\Rightarrow4-b=1\Rightarrow b=5\)

Với \(a=2;a=4\Rightarrow b\notin \mathbb{Z}\)

Vậy \(\left(a;b\right)=\left(1;5\right)\)

\(S^3=\left(\sqrt[3]{7+4\sqrt{3}+}\sqrt[3]{7-4\sqrt{3}}\right)^3\)

= \(7+4\sqrt{3}+7-4\sqrt{3}+3.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}.\left(a+b\right)\)

= 14+\(3.\sqrt{49-48}.S\)

= 14+3S

=> S³-3S=14+3S-3S=14

\(P=S^3-3S\)

\(P=\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)^3-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=7+4\sqrt{3}+3\left(\sqrt[3]{7+4\sqrt{3}}\right)^2.\sqrt[3]{7-4\sqrt{3}}+3.\sqrt[3]{7+4\sqrt{3}}\left(\sqrt[3]{7-4\sqrt{3}}\right)^2+7-4\sqrt{3}\text{​​}\text{​​}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=14+3\sqrt[3]{7+4\sqrt{3}}.\sqrt[3]{7-4\sqrt{3}}\text{​​}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{​​}\text{​​}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=14+3\sqrt[3]{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\text{​​}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{​​}\text{​​}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=14+3\sqrt[3]{49-48}\text{​​}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{​​}\text{​​}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=14+3\text{​​}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{​​}\text{​​}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)

\(P=14\)

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

\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)

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

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

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

\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)​

\(\sqrt{53-20\sqrt{7}}=a+b\sqrt{7}\)

\(\Leftrightarrow a+b\sqrt{7}=-5+2\sqrt{7}\)

=> a=-5; b=2

a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)

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

\(=\dfrac{10+2\sqrt{21}+10-2\sqrt{21}}{4}=\dfrac{20}{4}=5\)

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

\(=7\sqrt{3}-2\sqrt{2}-1\)