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\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
\(\sqrt{3-\sqrt{5}}\sqrt{3-\sqrt{5}}\)\(\sqrt{3+\sqrt{5}}\)\(+\sqrt{3+\sqrt{5}}\sqrt{3+\sqrt{5}}\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3-\sqrt{5}}\cdot\sqrt{3^2-5}+\sqrt{3+\sqrt{5}}\cdot\sqrt{3^2-5}\)=\(2\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)=\sqrt{2}\left(\sqrt{2\cdot3-2\sqrt{5}}+\sqrt{2\cdot3+2\sqrt{5}}\right)\) =\(=\sqrt{2}\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
b tuong tu nha ban ^.^
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
1) \(P=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2}\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2+\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+2\sqrt{60}+6\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+4\sqrt{15}+6\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(16+4\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{4\left(4+\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{4\left(16-15\right)}\)
\(=\sqrt{4\cdot1}\)
\(=\sqrt{4}\)
\(=2\)
2) \(Q=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(3-\sqrt{5}\right)^2}\sqrt{3+\sqrt{5}}+\sqrt{\left(3+\sqrt{5}\right)^2}\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(3-\sqrt{5}\right)^2\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(9-6\sqrt{5}+5\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(14-6\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{42+14\sqrt{5}-18\sqrt{5}-30}+\sqrt{42-14\sqrt{5}+18\sqrt{5}-30}\)
\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)