1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)

2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)

\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)

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

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

\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)

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

\(=\sqrt{5}-2-\sqrt{5}-2=-4\)

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=19-9=10\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5}-2-\sqrt{5}-2=-4\)

\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)

\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)

từ (1) and (2)

=>19A>19B

=>A>B

\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)

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

\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)

\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

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

\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)

Trả lời gấp giúp mk vs nka 

Bạn xem lại đầu bài đi chứ đấy là tỉ số phần trăm rồi mà

8%= 0,08 ; 25%= 0,25 ; 19%= 0,19 ; 12% = 0,12 ; 40%= 0,4

CHÚC BẠN HỌC TỐT

\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}\)

Với : 

\(\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}\)

Mặt khác : 

\(\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

Nên : 

\(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

Cách khác:

Ta có: \(\left(\sqrt{31}-\sqrt{19}\right)^2=50-2\sqrt{589}\)

\(\left(6-\sqrt{17}\right)^2=53-12\sqrt{17}=50+3-12\sqrt{17}\)

mà \(-2\sqrt{589}< 3-12\sqrt{17}\)

nên \(\sqrt{31}-\sqrt{19}>6-\sqrt{17}\)

$#Shả$

`A)=5/17xx(15/34+19/34)`

`=5/17xx1=5/17`

`B)=(13/12+7/12)+1/3`

`=5/3+1/3=2`

dấu . là nhân hay là phần ngăn cách ở hàng phần nghìn thế

a) \(28400:\left(100\times311-310\times100\right)\)
\(=28400:\left[\left(311-310\right)\times100\right]\)
\(=28400:100\)
\(=284\)
b) \(\dfrac{5}{11}+\dfrac{19}{15}+\dfrac{6}{11}+\dfrac{11}{16}+\dfrac{6}{15}+\dfrac{5}{16}\)
\(=\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+\left(\dfrac{19}{15}+\dfrac{6}{15}\right)+\left(\dfrac{11}{16}+\dfrac{5}{16}\right)\)
\(=1+\dfrac{5}{3}+1\)
\(=\dfrac{8}{3}+1\)
\(=\dfrac{11}{3}\)

a) 28.400 : ( 100 x 311 - 310 x 100 )     

= 28.400 :  [ 100 x ( 311 - 310  ]

= 28.400 : 100 x 1

= 0.284 x 1

= 0.284

b) \(\dfrac{5}{11}+\dfrac{19}{15}+\dfrac{6}{11}+\dfrac{11}{16}+\dfrac{6}{15}+\dfrac{5}{16}\)

\(=\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+\left(\dfrac{19}{15}+\dfrac{6}{15}\right)+\left(\dfrac{11}{16}+\dfrac{5}{16}\right)\)

\(=1+\dfrac{5}{3}+1\)

\(=\dfrac{11}{3}\)