\(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
1)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
2)\(\sqrt{35+12\sqrt{6}}-\sqrt{35-12\sqrt{6}}\)
3)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
`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`
Tính
a) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
b)\(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
Tính
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{4+\sqrt{7}} -\sqrt{4-\sqrt{7}}\)
c) \(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
a: =2-căn 3-2-căn 3
=-2căn 3
b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)
\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)
=>\(A=\sqrt{10+2\sqrt{5}}\)
Tính:
A=\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
B=\(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
C=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D=\(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)(2 cách)
F=\(\dfrac{\sqrt{17-12\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}-\dfrac{\sqrt{17}+12\sqrt{2}}{\sqrt{3+2\sqrt{2}}}\)
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
Cách 1:
\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)
\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
Cách 2:
\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)
\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)
Vì $E>0$ nên $E=2$
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(8+2\sqrt{16-10-2\sqrt{5}}\)
\(8+2\sqrt{6-2\sqrt{5}}\)
\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left(\sqrt{5}-1\right)\)
\(8+2\sqrt{5}-2\)
\(6+2\sqrt{5}\)
\(\left(\sqrt{5+1}\right)^2\)
\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(|\sqrt{5}+1|\)
\(\sqrt{5}+1\)
\(\text{thư ngu như chó}\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=>\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)
=>\(A^2=8+2\cdot\sqrt{6-2\sqrt{5}}\)
=>\(A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
=>\(A=\sqrt{5}+1\)
Rút gọn biểu thức:
a)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
b)\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
c)\(5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}\)
d)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
e)\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
Rút gọn : \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Cho \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
B2 = \(4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
= \(8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}\)
= \(8+2\sqrt{5-2\sqrt{5}+1}\)
= \(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
= \(8+2.\left(\sqrt{5}-1\right)\) (do \(\sqrt{5}>1\))
= \(6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1\)
= \(\left(\sqrt{5}+1\right)^2\)
=> B = \(\sqrt{5}+1\)
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}\right)^2+\left(\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2+2\sqrt{4+\sqrt{10+2\sqrt{5}}}\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.1+1^2}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left|\sqrt{5}-1\right|=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.1+1^2\)
\(=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\left(A>0\right)\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\) rút gọn
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ A=\sqrt{5}+1\)