Tính:
C = \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D = \(\sqrt{4-\sqrt{7}}+\sqrt{2-\sqrt{3}}\)
K
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
20 tháng 9 2021
a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)
c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)
d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
---------------------
$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
H
30 tháng 9 2023
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
__
C làm giống câu a, nhé.
__
\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
__
\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
18 tháng 6 2021
`a)\sqrt{9-4sqrt5}-sqrt5`
`=sqrt{5-2.2sqrt5+4}-sqrt5`
`=sqrt{(sqrt5-2)^2}-sqrt5`
`=|\sqrt5-2|-sqrt5`
`=sqrt5-2-sqrt5=-2`
`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`
`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`
`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`
`=|2-sqrt3|+|sqrt3-1|`
`=2-sqrt3+sqrt3-1=1`
`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`
`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`
`=sqrtx+7`
`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`
`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`
`=sqrt3+1-2sqrt3-1=-sqrt3`
`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)
1 tháng 8 2023
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: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)
\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)
=>\(A=\sqrt{10+2\sqrt{5}}\)
24 tháng 8 2021
`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`
25 tháng 8 2021
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: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
MH
24 tháng 9 2021
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
13 tháng 5 2022
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
\(C.\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\sqrt{7}-1-\sqrt{7}-1=-2\)
=> \(C=-\sqrt{2}\)
\(D.\sqrt{2}=\sqrt{8-2\sqrt{7}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{7}-1+\sqrt{3}-1\)
=> \(D=\frac{\sqrt{7}+\sqrt{3}-2}{\sqrt{2}}\)
\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(\Rightarrow\sqrt{2}.C=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
\(\Rightarrow C=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
\(D=\sqrt{4-\sqrt{7}}+\sqrt{2-\sqrt{3}}\)
\(\Rightarrow\sqrt{2}.D=\sqrt{2}\sqrt{4-\sqrt{7}}+\sqrt{2}\sqrt{2-\sqrt{3}}\)
\(=\sqrt{8-2\sqrt{7}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{7-2\sqrt{7}+1}+\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{7}-1+\sqrt{3}-1\)
\(=\sqrt{7}+\sqrt{3}-2\)
\(\Rightarrow D=\frac{\sqrt{7}+\sqrt{3}-2}{\sqrt{2}}=\frac{\sqrt{14}+\sqrt{6}-2\sqrt{2}}{2}\)