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B
23 tháng 7 2019
\(a,\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right).\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(=2\)
\(b,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
KQ
1
25 tháng 8 2019
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}.1}+1}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\left|\sqrt{3}+1\right|}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
=\(\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
=\(\sqrt{6+2.\left(\sqrt{2}.\sqrt{2-\sqrt{3}}\right)}\)
=\(\sqrt{6+2.\left(\sqrt{4-2\sqrt{3}}\right)}\)
=\(\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{6+2.\left|\sqrt{3}-1\right|}\)
=\(\sqrt{6+2\sqrt{3}-2}\)
=\(\sqrt{4+2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)
=\(\left|\sqrt{3}+1\right|\)
=\(\sqrt{3}+1\)
HP
6 tháng 8 2021
a, Sửa đề:
\(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}-\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{2\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{2\sqrt{6-3\sqrt{3}}}{3}\)
23 tháng 9 2015
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 \(a\cdot b=\sqrt{2}.\)
HN
17 tháng 11 2017
Cái này mà là toán lớp 9 đó hả. Giống toán rút gọn của lớp 7 thế.
\(A=\dfrac{7-6\sqrt{2}}{\sqrt{6}-\sqrt{3}-\sqrt{2}}=\dfrac{\left(6-6\sqrt{2}+3\right)-2}{\sqrt{6}-\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{6}-\sqrt{3}\right)^2-2}{\sqrt{6}-\sqrt{3}-\sqrt{2}}=\dfrac{\left(\sqrt{6}-\sqrt{3}-\sqrt{2}\right)\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}{\sqrt{6}-\sqrt{3}-\sqrt{2}}=\sqrt{6}-\sqrt{3}+\sqrt{2}\)
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) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
\(A=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(A=\sqrt{2-2\sqrt{2}.1+1}-\sqrt{4+2.2\sqrt{2}+2}\)
\(A=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(A=\left|\sqrt{2-1}\right|-\left|2+\sqrt{2}\right|\)
\(A=\sqrt{2}-1-2-\sqrt{2}\)|
\(A=-3\)
\(A=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{2}-1-\left(2+\sqrt{2}\right)\)
\(=\sqrt{2}-1-2-\sqrt{2}\)
\(=-1-2\)
\(=-3\)