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14 tháng 4 2021

a) \(\sqrt{16}\).\(\sqrt{25}\)+\(\sqrt{196}\):\(\sqrt{49}\)

=4.5+14/7

=20+2

=22

21 tháng 5 2021

a) \(\sqrt{16}\).\(\sqrt{25}\) + \(\sqrt{196}\) : \(\sqrt{49}\) = 4.5+14:9=22

b) 36:\(\sqrt{2.3^2.18}\) - \(\sqrt{169}\)= 36 :  \(\)18 - 13 = -11

c) \(\sqrt{\sqrt{81}}\) = 3

d) \(\sqrt{3^2+4^2}\)\(\sqrt{25}\)=5

a: \(=4\cdot5+14:7\)

=20+2

=22

31 tháng 3 2017

a) \(\sqrt{16}\cdot\sqrt{25}+\sqrt{196}:\sqrt{49}\)

\(=\sqrt{16\cdot25}+\sqrt{196:49}\)

\(=20+2=22\)

b) \(36:\sqrt{2\cdot3^2\cdot18}-\sqrt{169}\)

\(=36:\sqrt{324}-\sqrt{169}\)

\(=36:18-13=2-13=-11\)

c) \(\sqrt{\sqrt{81}}\)

\(=\sqrt{9}=3\)

d) \(\sqrt{3^2+4^2}\)

\(=\sqrt{9+16}=\sqrt{25}=5\)

7 tháng 6 2017

a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)

\(=4.5+14:7\)

\(=20+2=22\)

b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)

\(=36:18-13=-11\)

c) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)

d) \(\sqrt{3^2+4^2}=\sqrt{25}=5\)

26 tháng 10 2018

Bài làm:

Bài 1:

a)\(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)

= 4.5 + 14 : 7

= 20 + 2

= 22

b)\(36:\sqrt{2.3^2.18}-\sqrt{169}\)

= 36 : 18 - 14

= 2 - 14

= - 12

c)\(\sqrt{\sqrt{81}}\) = \(\sqrt{9}\) = 3

d)\(\sqrt{3^2+4^2}\)

= \(\sqrt{9+16}\)

= \(\sqrt{25}\)

= 5

26 tháng 10 2018

Làm sai rồi

14 tháng 4 2021

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

b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0

Bình phương 2 vế, ta có:

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

21 tháng 5 2021

a)  \(\left(\sqrt{3}-1\right)^2\)=\(\left(\sqrt{3}\right)^2\)- 2\(\sqrt{3}\) +1= 3- 2\(\sqrt{3}\) +1=4-2\(\sqrt{3}\)

b)  \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)\(|\sqrt{3}-1|\)-\(\sqrt{3}\)=\(\sqrt{3}\)-1-\(\sqrt{3}\)=-1

 

15 tháng 7 2017

a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)

14 tháng 5 2021

a) \(\dfrac{40}{27}\)

b) \(\dfrac{196}{45}\)

c) \(\dfrac{56}{9}\)

d) 1296

19 tháng 5 2021

a) \sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}

=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}

=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}

=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}.

b) \sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}

=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}

=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}

=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}

=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}.

c) \dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}

=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}

=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}

=\dfrac{56}{9}.

d) \sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}

=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}

=\sqrt{216.81 .(11+5)(11-5)}

=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}

=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296.

19 tháng 4 2021

a, Ta có  \(\sqrt{25-16}=\sqrt{9}=3\)

\(\sqrt{25}-\sqrt{16}=5-4=1\)

Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

13 tháng 5 2021

a) căn 25 - 16  > căn 25 - căn 16

 

b)Với a>b>0 nên  \sqrt{a},\sqrt{b},\sqrt{a-b} đều xác định

 

Để so sánh \sqrt{a}-\sqrt{b} và \sqrt{a-b} ta quy về so sánh \sqrt{a} và \sqrt{a-b}+\sqrt{b}.

 

+) (\sqrt{a})^2=a.

                                       

+) (\sqrt{a-b}+\sqrt{b})^2=(\sqrt{a-b})^2+2\sqrt{a-b}.\sqrt{b}+(\sqrt{b})^2=a-b+b+2\sqrt{a-b}.\sqrt{b}=a+2\sqrt{a-b}.\sqrt{b}

.

Do a>b>0 nên 2\sqrt{a-b}.\sqrt{b}>0

 

 

\Rightarrow a+2\sqrt{a-b}.\sqrt{b}>a

 

\Rightarrow (\sqrt{a-b}+\sqrt{b})^2>(\sqrt{a})^2

 

Do \sqrt{a},\sqrt{a-b}+\sqrt{b}>0 

 

\Rightarrow \sqrt{a-b}+\sqrt{b}>\sqrt{a}

 

\Leftrightarrow \sqrt{a-b}>\sqrt{a}-\sqrt{b} (đpcm)

 

Vậy \sqrt{a-b}>\sqrt{a}-\sqrt{b}.