câu cuối phải là \(-\sqrt{7}+3\) chứ nhỉ=)))

a)

\(A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)

\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\) (vì \(2-\sqrt{5}< 0;2\sqrt{2}-\sqrt{5}>0\) )

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

b)

\(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)

\(=2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\) (vì \(\sqrt{7}-2\sqrt{2}< 0;3-2\sqrt{2}>0\) )

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

`(5x-1)^3`

`=(5x)^3 -3*(5x)^2 *1+3*5x*1^2 -1^3`

`=125x^3 -75x^2 +15x-1`

`n_(S)=m/M=(3,2)/32=0,1(mol)`

`n_(O_2)=m/M=(6,4)/32=0,2(mol)`

\(PTHH:S+O_2-^{t^o}>SO_2\)

tỉ lệ          1 :      1     ;        1

n(mol)     0,1---->0,1----->0,1

\(\dfrac{n_S}{1}< \dfrac{n_{O_2}}{1}\left(\dfrac{0,1}{1}< \dfrac{0,2}{1}\right)\)

`=>S` hết, `O_2` dư, tính theo `S`

\(m_{SO_2}=n\cdot M=0,1\cdot64=6,4\left(g\right)\)

`(5-x^3 )(5+x^2 )`

`=25+5x^2 -5x^3 -x^5`

`1024:2^6 +160:(3^3 +53)-9^100 :9^99`

`=2^10 :2^6 +160:(27+53)-9^(100-99)`

`=2^(10-6)+160:80-9^1`

`=2^4 +2-9`

`=16+2-9`

`=18-9`

`=9`

5)

\(\dfrac{4}{\sqrt{5}-3}-\dfrac{1}{\sqrt{5}+2}\\ =\dfrac{4\left(\sqrt{5}+3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}-\dfrac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{4\sqrt{5}+12}{5-9}-\dfrac{\sqrt{5}-2}{5-4}\\ =\dfrac{4\sqrt{5}+12}{-4}-\left(\sqrt{5}-2\right)\\ =\dfrac{4\left(\sqrt{5}+3\right)}{-4}-\sqrt{5}+2\\ =-\left(\sqrt{5}+3\right)-\sqrt{5}+2\\ =-\sqrt{5}-3-\sqrt{5}+2\\ =-2\sqrt{5}-1\)

6)

\(\dfrac{1}{3-2\sqrt{2}}+\dfrac{2\sqrt{3}}{2+\sqrt{5}}\\ =\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{2\sqrt{3}\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\)

\(=\dfrac{3+2\sqrt{2}}{9-8}+\dfrac{4\sqrt{3}-2\sqrt{15}}{4-5}\\ =3+2\sqrt{2}-\dfrac{4\sqrt{3}-2\sqrt{15}}{5-4}\\ =3+2\sqrt{2}-\left(4\sqrt{3}-2\sqrt{15}\right)\\ =3+2\sqrt{2}-4\sqrt{3}+2\sqrt{15}\)