RÚT GỌN : \(2\sqrt{2+\sqrt{3}}\)
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HA
2
16 tháng 6 2021
TK:"https://www.google.com.vn/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjF-cL3pJzxAhWbPXAKHTjTAxwQFjABegQIBRAD&url=https%3A%2F%2Fhoidap247.com%2Fcau-hoi%2F996088&usg=AOvVaw3JxumatFPaPIuCWni48U22"
16 tháng 6 2021
`sqrt{3-2sqrt2}-sqrt{3+2sqrt2}`
`=sqrt{2-2sqrt2+1}-sqrt{2+2sqrt2+1}`
`=sqrt{(sqrt2-1)^2}-sqrt{(sqrt2+1)^2}`
`=sqrt2-1-sqrt2-1=-2`
L
1
21 tháng 8 2021
Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
V
1
AH
Akai Haruma
Giáo viên
15 tháng 5 2021
Lời giải:
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{3-(\sqrt{3}+1)}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
DT
8 tháng 6 2021
A = \(\sqrt{3+2\sqrt{2}}-\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{3+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2+2\sqrt{3}\left(\sqrt{2}+1\right)+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1+\sqrt{3}\right)^2}\)
= \(\left|\sqrt{2}+1\right|-\left|\sqrt{2}+\sqrt{3}+1\right|\)
= \(\sqrt{2}+1-\sqrt{2}-\sqrt{3}-1\)
= \(-\sqrt{3}\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
PN
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 7 2023
\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)
\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)
\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)
6 tháng 7 2021
a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=4-2\sqrt{4-3}=4-2=2\)
\(\Rightarrow A=-\sqrt{2}\)
b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)
\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow B=\sqrt{2}\)
24 tháng 6 2021
Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\sqrt{\dfrac{\left(1+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{2}}\)
= \(\dfrac{1+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\)
<=> A = 1
AT
20 tháng 6 2021
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2}+1-\sqrt{2}+1=2\)
\(2\sqrt{2+\sqrt{3}}=\sqrt{8+4\sqrt{3}}=\sqrt{8+2\sqrt{12}}=\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}=\sqrt{6}+\sqrt{2}\)