Tính:
a) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Mọi người giúp em gấp đi ạ!!!!!!!!!!!!!!!1
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PA
3 tháng 8 2017
\(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\times\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=2\sqrt{3+\sqrt{5}}\times\sqrt{2}\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\sqrt{6+2\sqrt{5}}\times\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}\times\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\left(\sqrt{5}+1\right)\times\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\left(5-1\right)\)
= 8
~ ~ ~
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=\left(2\sqrt{2}-\sqrt{5}\right)-\left(3\sqrt{5}+2\sqrt{2}\right)\)
\(=-4\sqrt{5}\)
3 tháng 8 2017
a. \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left[2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right]\left(\sqrt{10}-\sqrt{2}\right)=\left(2\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left(2\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left[2\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}\right]\left(\sqrt{10}-\sqrt{2}\right)=\left[2\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)\right]\left(\sqrt{10}-\sqrt{2}\right)=\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{10}-\sqrt{2}\right)=10-2=8\)
b. \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}=-4\sqrt{5}\)
8 tháng 7 2018
a) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+12\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{8}\right)^2-2.\sqrt{8}.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2.3\sqrt{5}.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=\left|\sqrt{8}-\sqrt{5}\right|-\left|3\sqrt{5}+2\sqrt{2}\right|\)
= √8 - √5 - 3√5 - 2√2 = -4√5
b) (1+√3-√2).(1+√3+√2)= [(1+√3)^2-(√2)^2] = 4+2√3-2=2+2√3
8 tháng 7 2018
a) =sprt{13-=sprt{160}} - =sprt{53+4=sprt{90}}
= =sprt{(=sprt{8} - =sprt{5})2 } - =sprt{(=sprt{45} + =sprt{8})2 }
= =sprt{8} - =sprt{5} - =sprt{45} - =sprt{8}
= -3=sprt{5}
b) ( 1 + =sprt{3} - =sprt{2} )( 1+ =sprt{3} + =sprt{2} )
= ( 1 + =sprt{3} )2 - (=sprt{2})2
= 4 + 2=sprt{3} - 2
=2 + 2=sprt{3}
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) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
Q
24 tháng 6 2017
a) \(\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}=\sqrt{7}-\sqrt{5}\)
b) \(\sqrt{4+\sqrt{15}}=...\)
c) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}=\left(3\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\\ =\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=9-2=7\)
d) \(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}=\left(\sqrt{7}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}\\ =\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=7-5=2\)
e) \(\sqrt{7-2\sqrt{10}-\sqrt{7+2\sqrt{10}}}=\sqrt{7-2\sqrt{10}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\\ =\sqrt{7-2\sqrt{10}-\left(\sqrt{2}+\sqrt{5}\right)}=\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\\ =\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\)
f) \(\sqrt{13-\sqrt{160}+\sqrt{53}+4\sqrt{90}}=\sqrt{13-4\sqrt{10}+\sqrt{53}+12\sqrt{10}}\\ =\sqrt{13+8\sqrt{10}+\sqrt{53}}\)
16 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
5 tháng 7 2018
Mình làm luôn nhé :
\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v
\(a.\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.2\sqrt{2}.3\sqrt{5}+8}=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}=-4\sqrt{5}\) \(b.\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}.\sqrt{3}+3}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)