Question

Đã trả lời bởi chuyên gia

Answer 1

a,

${(1 + 2 \sqrt{2})}^{2} = 1 + 4 \sqrt{2} + 8 = 9 + \sqrt{4^{2} . 2} = 9 + \sqrt{32} 4^{2} = 16 = 9 + 7 = 9 + \sqrt{7^{2}} = 9 + \sqrt{49} D o \sqrt{32} < \sqrt{49} \Rightarrow {(1 + 2 \sqrt{2})}^{2} < 4 \Rightarrow 1 + 2 \sqrt{2} < 4 b, {(2 \sqrt{6} - 1)}^{2} = 24 - 4 \sqrt{6} + 1 = 25 - \sqrt{4^{2} . 6} = 25 - \sqrt{96} 4^{2} = 16 = 25 - 9 = 25 - \sqrt{9^{2}} = 25 - \sqrt{81} D o - \sqrt{96} < - \sqrt{81} \Rightarrow {(2 \sqrt{6} - 1)}^{2} < 4^{2} \Rightarrow 2 \sqrt{6} - 1 < 4 c, {(\sqrt{3} - 2)}^{2} = 3 - 4 \sqrt{3} + 4 = 7 - \sqrt{4^{2} . 3} = 7 - \sqrt{48} {(\sqrt{0, 5})}^{2} = 0, 5 = 7 - 6, 5 = 7 - \sqrt{6, 5^{2}} = 7 - \sqrt{42, 25} D o - \sqrt{48} < - \sqrt{42, 25} \Rightarrow {(\sqrt{3} - 2)}^{2} < {(\sqrt{0, 5})}^{2} \Rightarrow \sqrt{3} - 2 < \sqrt{0, 5} d, - 3 \sqrt{3} = - \sqrt{3^{2} . 3} = - \sqrt{27} - 2 \sqrt{7} = - \sqrt{2^{2} . 7} = - \sqrt{28} D o \sqrt{27} < \sqrt{28} \Rightarrow - \sqrt{27} > - \sqrt{28} \Rightarrow - 3 \sqrt{3} > - 2 \sqrt{7}$

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Answer 2

a,

${(1 + 2 \sqrt{2})}^{2} = 1 + 4 \sqrt{2} + 8 = 9 + \sqrt{4^{2} . 2} = 9 + \sqrt{32} 4^{2} = 16 = 9 + 7 = 9 + \sqrt{7^{2}} = 9 + \sqrt{49} D o \sqrt{32} < \sqrt{49} \Rightarrow {(1 + 2 \sqrt{2})}^{2} < 4 \Rightarrow 1 + 2 \sqrt{2} < 4 b, {(2 \sqrt{6} - 1)}^{2} = 24 - 4 \sqrt{6} + 1 = 25 - \sqrt{4^{2} . 6} = 25 - \sqrt{96} 4^{2} = 16 = 25 - 9 = 25 - \sqrt{9^{2}} = 25 - \sqrt{81} D o - \sqrt{96} < - \sqrt{81} \Rightarrow {(2 \sqrt{6} - 1)}^{2} < 4^{2} \Rightarrow 2 \sqrt{6} - 1 < 4 c, {(\sqrt{3} - 2)}^{2} = 3 - 4 \sqrt{3} + 4 = 7 - \sqrt{4^{2} . 3} = 7 - \sqrt{48} {(\sqrt{0, 5})}^{2} = 0, 5 = 7 - 6, 5 = 7 - \sqrt{6, 5^{2}} = 7 - \sqrt{42, 25} D o - \sqrt{48} < - \sqrt{42, 25} \Rightarrow {(\sqrt{3} - 2)}^{2} < {(\sqrt{0, 5})}^{2} \Rightarrow \sqrt{3} - 2 < \sqrt{0, 5} d, - 3 \sqrt{3} = - \sqrt{3^{2} . 3} = - \sqrt{27} - 2 \sqrt{7} = - \sqrt{2^{2} . 7} = - \sqrt{28} D o \sqrt{27} < \sqrt{28} \Rightarrow - \sqrt{27} > - \sqrt{28} \Rightarrow - 3 \sqrt{3} > - 2 \sqrt{7}$

Answer 3

`c)` Ta có :

`\sqrt{0,5^2} = | 0,5 | = 0,5`

`(\sqrt{3})^2 - 2^2 = 3 - 4 = -1`

Mà `0,5 > -1`

`=> \sqrt{0,5} > \sqrt{3} - 2`

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`d)` Ta có :

`-3\sqrt{3} = -\sqrt{3^2 . 3} = -\sqrt{9.3} = -\sqrt{27}`

`-2\sqrt{7} = -\sqrt{2^2.7} = -\sqrt{4.7} = -\sqrt{28}`

Mà `-\sqrt{27} > -\sqrt{28}`

`=> -3\sqrt{3} > -2\sqrt{7} `

Answer 4

`a)`

`4` và `1+`$\sqrt{22}$

Ta có `4=1+` $\sqrt{9}$

Mà $\sqrt{22}$`>`$\sqrt{9}$

`=>1+` $\sqrt{22}$`>1+`$\sqrt{9}$

Hay `1+`$\sqrt{22}$`>4`

`b) `

`4` và `2`$\sqrt{6}$`-1`

Ta có `4=2`$\sqrt{4}$

Lại có `2`$\sqrt{6}$`>2`$\sqrt{4}$

Mà `2`$\sqrt{6}$`-1<2`$\sqrt{4}$

Vì ta thấy `2` $\sqrt{4}$`+0` vẫn bằng` 2`$\sqrt{4}$

Mà `0> -1` nên `=>` `4>` `2`$\sqrt{6}$`-1`