Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{5}{2^6}+\frac{10}{2^6}=\frac{15}{2^6}\)

Lại có B = \(\frac{11}{2^7}+\frac{9}{2^6}=\frac{5,5}{2^6}+\frac{9}{2^6}=\frac{14,5}{2^6}\)

Vì \(\frac{15}{2^6}>\frac{14,5}{2^6}\Rightarrow A>B\)

b) Ta có : \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\)

Lại có B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=\frac{-157}{10^{2006}}\)

Vì \(\frac{-85}{10^{2006}}>\frac{-157}{10^{2006}}\Rightarrow A< B\)

10)=5.382332347

11)=1.918230732

\(=\sqrt{10+2\cdot\sqrt{2}\cdot\sqrt{3}-2\cdot\sqrt{2}\cdot\sqrt{5}-2\cdot\sqrt{3}\cdot\sqrt{5}}\)

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

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

Phép tính:

\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{6}=1421411372\)

\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{3}+\sqrt{6}=5602951922\)

P/s: Em ko biết đúng hay sai đâu mới lớp 4 thôi à

Mình thấy đề cứ sai sai.

\(\dfrac{2\sqrt{15}-2\sqrt{10}-3+\sqrt{6}}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)

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

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

`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`

`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`

`<=>A^2=8+2sqrt{6-2sqrt5}`

`<=>A^2=8+2sqrt{(sqrt5-1)^2}`

`<=>A^2=8+2(sqrt5-1)`

`<=>A^2=6+2sqrt5=(sqrt5+1)^2`

`<=>A=sqrt5+1(do \ A>0)`

`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`

Vì `35+12sqrt6>35-12sqrt6`

`=>B>0`

`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`

`<=>B^2=70-2sqrt{361}`

`<=>B^2=70-2sqrt{19^2}=70-38=32`

`<=>B=sqrt{32}=4sqrt2(do \ B>0)`

`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`

`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`

`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`

`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`

`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`

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

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

`=5-3=2`

\(=\sqrt{2+3+5-2\cdot\sqrt{2}\cdot\sqrt{3}+2\sqrt{2}\cdot\sqrt{5}-2\cdot\sqrt{3}\cdot\sqrt{5}}\)

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