\(\sqrt{10-\sqrt{19}}-\sqrt{10+\sqrt{19}}\)

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\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

\(=\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

\(=\sqrt{100-19}\)

= \(\sqrt{81}\)

\(=9\)

4 tháng 7 2018

\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

=\(\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

=\(\sqrt{100-19}\)

=\(\sqrt{81}\)

= 9 (đpcm)

AH
Akai Haruma
Giáo viên
6 tháng 10 2019

a)

\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)

\(=2(4^2-15)=2\)

b)

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)

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

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

AH
Akai Haruma
Giáo viên
6 tháng 10 2019

c)

\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)

\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)

d)

\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)

\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)

Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)

\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)

11 tháng 8 2019

b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))

t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)

t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))

t = (\(\sqrt{5}\) -1)2.(3 +\(\sqrt{5}\))

t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))

t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)

t = 6

1 tháng 5 2017

Làm tới dòng thứ 3 máy đơ, 2 lần rồi T,T

Mình chia làm 2 phần tính nhé

\(A=\frac{4\sqrt{2}}{\sqrt{10-2\sqrt{21}}}+\frac{3}{\sqrt{15+6\sqrt{6}}}-\frac{1}{\sqrt{19-6\sqrt{10}}}\)

\(A=\frac{4\sqrt{2}}{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}+\frac{3}{\sqrt{\left(\sqrt{9}+\sqrt{6}\right)^2}}-\frac{1}{\sqrt{\left(\sqrt{10}-\sqrt{9}\right)^2}}\)

\(A=\frac{4\sqrt{2}}{\sqrt{7}-\sqrt{3}}+\frac{3}{3+\sqrt{6}}-\frac{1}{\sqrt{10}-3}\)

\(A=\frac{4\sqrt{2}\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{3\left(3-\sqrt{6}\right)}{9-6}-\frac{1\left(\sqrt{10}+3\right)}{10-9}\)

\(A=\frac{4\sqrt{14}+4\sqrt{6}}{4}+\frac{9-3\sqrt{6}}{3}-\sqrt{10}-3\)

\(A=\sqrt{14}+\sqrt{6}+3-\sqrt{6}-\sqrt{10}-3\)

\(A=\sqrt{14}-\sqrt{10}\)

\(B=\sqrt{6+\sqrt{35}}\)

\(B=\frac{\sqrt{2}\left(\sqrt{6+\sqrt{35}}\right)}{\sqrt{2}}\)

\(B=\frac{\sqrt{12+2\sqrt{35}}}{\sqrt{2}}\)

\(B=\frac{\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}}{\sqrt{2}}\)

\(B=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)

\(\Rightarrow M=A.B=\left(\sqrt{14}-\sqrt{10}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)

\(M=\sqrt{2}\left(\sqrt{7}-\sqrt{5}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)

\(M=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(M=\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2\)

\(M=7-5=2\)

28 tháng 6 2017

18-6\(\sqrt{10}\)1=8-360=-342

19+8\(\sqrt{3}\)=19+192=211

27+10\(\sqrt{2}\)=27+200=227

28+12\(\sqrt{2}\)=28+288=316

7 tháng 7 2016

bạn kiểm tra lại biểu thức A đi bạn

 

2 tháng 8 2017

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)

8 tháng 7 2016

Đặt \(a=\sqrt{x^2-6x+19},a\ge0\) ; \(b=\sqrt{x^2-6x+10},b\ge0\)

\(\Rightarrow\begin{cases}a-b=3\\a^2-b^2=9\end{cases}\)  \(\Rightarrow A=a+b=3\)