Rút gọn biểu thức:
\(a,\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(b,\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
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H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)
\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)
\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)
\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)
\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)
\(=7-\sqrt{21}+\sqrt{21}-3\)
\(=4\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
15 tháng 5 2021
`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
15 tháng 5 2021
`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`
`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`
`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`
`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`
`=2(\sqrt5-1)sqrt{6+2\sqrt5}`
`=2(\sqrt5-1)(\sqrt5+1)`
`=2(5-1)`
`=8`
`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=\sqrt2(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`
`=(4+\sqrt{15})(\sqrt5-\sqrt3)(\sqrt5-\sqrt3)`
`=(4+\sqrt{15})(8-2\sqrt{15})`
`=2(4+\sqrt{15})(4-\sqrt{15})`
`=2(16-15)`
`=2`
11 tháng 10 2017
Phần a sai đề sửa đề
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-{12\sqrt{5}}}}}\)
=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(2\sqrt{5}-3)^2 } } } \)
=\(\sqrt{5-\sqrt{3-2\sqrt{5}+3 }}\)
=\(\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2 } } \)
=\(\sqrt{\sqrt{5}-\sqrt{5}+1 } \)
=1
11 tháng 10 2017
B=\((\sqrt{4+\sqrt{15} }) \sqrt{2}(\sqrt{5}-\sqrt{3})(\sqrt{4-\sqrt{15} })({\sqrt{4+\sqrt{15} }) } \)
=(\((\sqrt{4+\sqrt{15} })\sqrt{2}(\sqrt{5}-\sqrt{3}) \)
=\((\sqrt{8+2\sqrt{15} })(\sqrt{5}-\sqrt{3}) \)
=\((\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) \)
=2
9 tháng 6 2019
a,A.√2= √(4+2√3)-√(4-2√3)
= √(1+√3)2 -√( √3 -1)2
= 1+√3-√3+1= 2
=> A= 2/√2=√2
9 tháng 6 2019
B2= (4+√15)2.(4-√15).(√10-√6)2
= (4+√15).1.(16-4√15)
= (4+√15).(4-√15).4
= 4
=> B = √4 = 2
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
1 tháng 7 2016
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 \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(A=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(B=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\)