#)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!

HẾ lu van buồi

vân buồi ơi kết bạn ko

a) \(\frac{2}{4-3\sqrt{2}}-\frac{2}{4+3\sqrt{2}}\)

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

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

\(=\frac{12\sqrt{2}}{-2}\)

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

b) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{4\sqrt{35}}{2}\)

\(=2\sqrt{35}\)

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

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

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

\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)

2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)

3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)

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

5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)

6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)

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

a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)

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

c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)

d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)