\(\sqrt{9-\sqrt{ }17}\) - \(\sqrt{9+\sqrt{ }17}\)
rút gọn K
K = \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)
\(K=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}-\sqrt{\left(-8\right)^2}\)
\(=\sqrt{81-17}-8=\sqrt{64}-8=8-8=0\)
Rút gọn \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)
CMR:
\(a)\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\\ b)2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
a) Ta có: \(VT=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\cdot\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)=VP(đpcm)
b) Ta có: \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
=9=VP(đpcm)
Tính:
a.\(\sqrt{4+\sqrt{7}}\) - \(\sqrt{4-\sqrt{7}}\)
b.\(\sqrt{4-\sqrt{15}}\) - \(\sqrt{4+\sqrt{15}}\)
c.\(\sqrt{2+\sqrt{3}}\) + \(\sqrt{2-\sqrt{3}}\)
d.\(\sqrt{9+\sqrt{17}}\) - \(\sqrt{9-\sqrt{17}}\)
Mong mn giúp em bài này ạ .Em đang cần gấp !!
`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`
`=2/sqrt2=sqrt2`
`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`
`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`
`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`
`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`
`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`
`=(-2sqrt3)/sqrt2=-sqrt6`
`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`
`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`
`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`
`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`
`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`
`=(2sqrt3)/sqrt2=sqrt6`
`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`
`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`
`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`
`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`
`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`
`=2/sqrt2=sqrt2`
a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)
b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
Tính:
\(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
Ta có: \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)
rút gọn
C=\(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}\)
Đặt \(A=\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)
\(\Leftrightarrow A^2=18-2\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\)
\(=18-2\sqrt{81-17}=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow C=A-\sqrt{2}=0\)
chứng minh:
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\)
Đặt \(A=\left(\sqrt{9-\sqrt{17}}\right).\left(\sqrt{9+\sqrt{17}}\right)\)
Ta có: \(A^2=\left[\left(\sqrt{9-\sqrt{17}}\right).\left(\sqrt{9+\sqrt{17}}\right)\right]=\left(9-\sqrt{17}\right).\left(9+\sqrt{17}\right)\)
\(=9^2-\left(\sqrt{17}\right)^2=81-17=64\)
\(=>A=\sqrt{64}=8\)
Xét vế trái:
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right)^2}.\sqrt{\left(\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)
\(=\left|\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right|.\left|\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right|\)
\(=\left(\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right).\left(\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right)\)
\(=\frac{17}{2}+\frac{\sqrt{17}}{2}-\frac{\sqrt{17}}{2}-\frac{1}{2}\)
\(=\frac{17}{2}-\frac{1}{2}=8\)
Vậy: \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8.\)
(Nhớ k cho mình với nha!)
Rút gọn
H= \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)
K=\(\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right).\sqrt{5}\)
a) \(H=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)
b) \(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right).\sqrt{5}\)
\(=\sqrt{20}.\sqrt{5}-3\sqrt{5}.\sqrt{5}+\sqrt{80}.\sqrt{5}\)
\(=\sqrt{100}-3.5+\sqrt{400}=\sqrt{10^2}-15+\sqrt{20^2}\)
\(=10-15+20=15\)
\(H=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{9^2-\left(\sqrt{17}\right)^2}\)
\(=\sqrt{81-17}\)
\(=\sqrt{64}=8\)
\(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right)\cdot\sqrt{5}\)
\(=\sqrt{20}\cdot\sqrt{5}-3\sqrt{5}\cdot\sqrt{5}+\sqrt{80}\cdot\sqrt{5}\)
\(=\sqrt{20\cdot5}-3\sqrt{5\cdot5}+\sqrt{80\cdot5}\)
\(=\sqrt{100}-3\sqrt{25}+\sqrt{400}\)
\(=10-3\cdot5+20\)
\(=15\)
\(H=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\)
\(K=\left(2\sqrt{5}-3\sqrt{5}+4\sqrt{5}\right)\sqrt{5}=3\sqrt{5}.\sqrt{5}=3.5=15\)
Đây là câu trả lời của mình, bạn chỉ cần áp dụng kĩ năng tính toán cơ bản là ra, học kĩ kiến thức cơ bản nhé
Chứng minh :
a) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\)
b) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
a) \(VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
=\(\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}=\sqrt{64}=8=VP\)
b) \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)