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2 tháng 9 2017
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
= \(\sqrt{3.4-3\sqrt{7}}-\sqrt{3.4+3\sqrt{7}}\)
= \(\sqrt{3.\left(4-\sqrt{7}\right)}-\sqrt{3.\left(4+\sqrt{7}\right)}\)
= \(\sqrt{3}.\sqrt{4-\sqrt{7}}-\sqrt{3}.\sqrt{4+\sqrt{7}}\)
= \(\sqrt{3}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)\)
\(\)≈ \(-2,449\)
2 tháng 9 2017
\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)
= \(\sqrt{\dfrac{13}{4}+\dfrac{4\sqrt{3}}{4}}-\sqrt{\dfrac{7}{4}-\dfrac{4\sqrt{3}}{4}}\)
= \(\sqrt{\dfrac{13+4\sqrt{3}}{4}}-\sqrt{\dfrac{7-4\sqrt{3}}{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}}{\sqrt{4}}-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
≈ \(2,098\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 3 2019
a/ \(2\sqrt{10}-10\sqrt{10}+9\sqrt{10}=\sqrt{10}\)
b/ \(\frac{-1\left(4-3\sqrt{2}\right)+1\left(4+3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\frac{-4+3\sqrt{2}+4+3\sqrt{2}}{16-18}=\frac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c/ \(\left(3+\sqrt{5}\right).\sqrt{2}.\sqrt{7-3\sqrt{5}}=\left(3+\sqrt{5}\right)\sqrt{14-6\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
d/ \(3\sqrt{2}-4\sqrt{2}+5\sqrt{2}=4\sqrt{2}\)
e/ \(\sqrt{19+8\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=4+\sqrt{3}+2-\sqrt{3}=6\)
Y
17 tháng 6 2019
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+\sqrt{12}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\frac{4+\sqrt{4^2-7}}{2}}+\sqrt{\frac{4-\sqrt{4^2-7}}{2}}-\left(\sqrt{\frac{4+\sqrt{4^2-7}}{2}}-\sqrt{\frac{4-\sqrt{4^2-7}}{2}}\right)+\left(\sqrt{3}+1\right)^2\)
( áp dụng công thức căn phức tạp )
\(=2\sqrt{\frac{4-3}{2}}+4+2\sqrt{3}\)
\(=\sqrt{2}+4+2\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 6 2019
\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)
\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)^2=\sqrt{2}+4+2\sqrt{3}\)
NC
25 tháng 8 2020
1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}-10-5\sqrt{10}\)
\(=-10\)
2) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=14-2\sqrt{21}-7+2\sqrt{21}\)
\(=7\)
3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)
QP
25 tháng 10 2016
1,=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{12}+4}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
=\(\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{4+2\sqrt{3}}\)
=\(\sqrt{3}+1\)
16 tháng 7 2018
\(a.\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3=11-4\sqrt{6}\)
\(b.\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-2=4+2\sqrt{3}-2=2+2\sqrt{3}\) \(c.\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}=6+4=10\) \(d.\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\sqrt{11-7}=2\sqrt{11}-4\) \(e.\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\sqrt{2}\) \(f.\sqrt{21-12\sqrt{3}}-\sqrt{3}=\sqrt{12-2.2\sqrt{3}.3+9}-\sqrt{3}=2\sqrt{3}-3-\sqrt{3}=\sqrt{3}-3\)
\(g.\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{3+2\sqrt{3}+1}=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(3-4\right)=-2\)
\(h.\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-2.4\sqrt{2}+2}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
7 tháng 8 2022
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
WR
23 tháng 6 2019
a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=1+\sqrt{3}\)
b) Tương tự a) đ/s =5
=\(\sqrt[3]{64.2}+\sqrt[3]{-125.2}-7\sqrt[3]{8.2}\)
= \(4\sqrt[3]{2}-5\sqrt[3]{2}-14\sqrt[3]{2}=-15\sqrt[3]{2}\)