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17 tháng 8 2021

\(\sqrt{15}-1< \sqrt{10}\)

17 tháng 8 2021

cám ơn

13 tháng 8 2018

a) Có 7 = 3 + 4 = \(\sqrt{9}+\sqrt{16}\)

mà 7 < 9 => \(\sqrt{7}< \sqrt{9}\)

15 < 16 => \(\sqrt{15}< \sqrt{16}\)

=> \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

=> \(\sqrt{7}+\sqrt{15}< 7\)

Vậy \(\sqrt{7}+\sqrt{15}< 7\)

b) Có 21 > 20

=> \(\sqrt{21}>\sqrt{20}\)

=> \(\sqrt{21}-\sqrt{6}>\sqrt{20}-\sqrt{6}\) (1)

Lại có 5 < 6

=> \(\sqrt{5}< \sqrt{6}\)

=> \(-\sqrt{5}>-\sqrt{6}\)

=> \(\sqrt{21}-\sqrt{5}>\sqrt{21}-\sqrt{6}\) (2)

Từ (1) và (2) => \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

Vậy \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

c) Có 27 > 25 => \(\sqrt{27}>\sqrt{25}\)

6 > 4 => \(\sqrt{6}>\sqrt{4}\)

=> \(\sqrt{27}+\sqrt{6}\) > \(\sqrt{25}+\sqrt{4}\)

=> \(\sqrt{27}+\sqrt{6}\) > 5 + 2

= >\(\sqrt{27}+\sqrt{6}+1>5+2+1\)

=> \(\sqrt{27}+\sqrt{6}+1>8\)

=> \(\sqrt{27}+\sqrt{6}+1>7\) (vì 8 > 7) (1)

Lại có 49 > 48

=> \(\sqrt{49}>\sqrt{48}\)

=> 7 > \(\sqrt{48}\) (2)

Từ (1) và (2) => \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)

Vậy \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)


22 tháng 6 2016

\(\sqrt{2}+1>\sqrt{1}+1=2\) 2

xong rui ban

22 tháng 6 2016

A) \(2\sqrt{31}=\sqrt{4\cdot31}=\sqrt{124}>\sqrt{10}\)

B)\(\sqrt{3-1}=\sqrt{2}>\sqrt{1}=1\)

C) \(-3\sqrt{11}=-\sqrt{99}>-\sqrt{144}=-12\)

14 tháng 8 2023

\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)

\(=\dfrac{\sqrt{5}}{2}\)

14 tháng 8 2023

\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)

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

\(=-\dfrac{\sqrt{21}}{7}\)

____________

\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{10}}{2}\)

26 tháng 6 2021

`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`

25 tháng 6 2023

\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)

25 tháng 6 2023

\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)

\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)

2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)

13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

\(=-\sqrt{5}\)

14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

=2

29 tháng 8 2021

12.

\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)

13.

\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

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

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

\(=-\sqrt{5}\)