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\(a=\sqrt{2+\sqrt{2}}.\sqrt{3+\sqrt{7+\sqrt{2}}}\)
\(b=\sqrt{3+\sqrt{6+\sqrt{7+\sqrt{2}}}}.\sqrt{3-\sqrt{6+\sqrt{7+\sqrt{2}}}}\)
tinh a.b
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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) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
25 tháng 10 2023
a: \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\left|\sqrt{2}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)
\(=-\sqrt{7}+\sqrt{7}-1+6=5\)
12 tháng 7 2021
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
20 tháng 8 2020
Ta có :
\(b^2=\left(3+\sqrt{6+\sqrt{7+\sqrt{2}}}\right)\left(3-\sqrt{6+\sqrt{7+\sqrt{2}}}\right)\)
\(b^2=9-\left(6+\sqrt{7+\sqrt{2}}\right)\)
\(b^2=3-\sqrt{7+\sqrt{2}}\)
\(\Rightarrow b=\sqrt{3-\sqrt{7+\sqrt{2}}}\)
Tích ab :
\(ab=\sqrt{2+\sqrt{2}}.\sqrt{3+\sqrt{7+\sqrt{2}}}.\sqrt{3-\sqrt{7+\sqrt{2}}}\)
\(ab=\sqrt{2+\sqrt{2}}.\left(9-7-\sqrt{2}\right)\)
\(ab=\sqrt{2+\sqrt{2}}.\left(2-\sqrt{2}\right)\)
P/s : làm được thế này thui . Sai bỏ qua
12 tháng 10 2023
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
MH
12 tháng 10 2023
a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
7 tháng 10 2022
a: \(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=1\)
b: \(=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{9-6-\sqrt{7+\sqrt{2}}}\)
\(=\sqrt{2+\sqrt{2}}\cdot\sqrt{9-7-\sqrt{2}}\)
\(=\sqrt{2}\)
Áp dụng hằng đẳng thức \(\left(x-y\right)\left(x+y\right)=x^2-y^2\) và tính chất \(\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}\)ta nhận được
\(b=\sqrt{3+\sqrt{6+\sqrt{7+\sqrt{2}}}}\cdot\sqrt{3-\sqrt{6+\sqrt{7+\sqrt{2}}}}\)
\(=\sqrt{\left(3+\sqrt{6+\sqrt{7+\sqrt{2}}}\right)\left(3-\sqrt{6+\sqrt{7+\sqrt{2}}}\right)}\)
\(=\sqrt{3^2-\left(6+\sqrt{7+\sqrt{2}}\right)}=\sqrt{3-\sqrt{7+\sqrt{2}}.}\)
Do đó \(b=\sqrt{3-\sqrt{7+\sqrt{2}}}.\) Suy ra
\(a\cdot b=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{3-\sqrt{7+\sqrt{2}}}\)
\(=\sqrt{2+\sqrt{2}}\sqrt{\left(3+\sqrt{7+\sqrt{2}}\right)\left(3-\sqrt{7+\sqrt{2}}\right)}\)
\(=\sqrt{2+\sqrt{2}}\sqrt{3^2-\left(7+\sqrt{2}\right)}\)
\(=\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2^2-2}=\sqrt{2}.\)
Vậy \(a\cdot b=\sqrt{2}.\)