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a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)
a) \(\left(\sqrt{8}+\sqrt{72}-\sqrt{2}\right).\sqrt{2}\)
\(=\left(2\sqrt{2}+6\sqrt{2}-\sqrt{2}\right).\sqrt{2}\)
\(=7\sqrt{2}.\sqrt{2}=7.2=14\)
b) \(\left(\sqrt{5}+\sqrt{2}+1\right)\left(\sqrt{5}-1\right)\)
\(=5-\sqrt{5}+\sqrt{10}-\sqrt{2}+\sqrt{5}-1\)
\(=4+\sqrt{10}-\sqrt{2}\)
c) \(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)
\(=\left(\sqrt{4+\sqrt{7}}\right)^2-2\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}+\left(\sqrt{4-\sqrt{7}}\right)^2\)
\(=\left(4+\sqrt{7}\right)-6+\left(4-\sqrt{7}\right)\)
\(=4+\sqrt{7}-6+4-\sqrt{7}=2\)
d) \(\left(\sqrt{2}+1+\sqrt{3}\right).\left(\sqrt{2}+1-\sqrt{3}\right)\)
\(=\left(\sqrt{2}+1\right)^2-3=2+2\sqrt{2}+1-3=2\sqrt{2}\)
e) \(\left(\sqrt{\frac{9}{2}}+\sqrt{\frac{1}{2}}-\sqrt{2}\right).\sqrt{2}\)
\(=3+1-2=2\)(nhân vào)
f) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
\(=\left(\sqrt{75}+\sqrt{45}\right):\sqrt{15}=\sqrt{5}+\sqrt{3}\)(chia đa tức cho đơn thức)
có sai xót mong m.n bỏ qa cho ♥
1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)
4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)
4A = \(8\sqrt{2}+1\)
⇒ A = \(\frac{\text{}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)
2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)
4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)
4B = \(12\sqrt{2}+1\)
⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)
3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)
= \(\sqrt{2-\sqrt{3}}\). \(\sqrt{2}.\left(\sqrt{3}+1\right)\)
= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}\right)^2\) - 12
= 3 - 1
= 2
4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)
= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)
= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)
= \(\sqrt{7}\) . (7 - 3)
= 4\(\sqrt{7}\)
5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)
= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)
= 10 - 2
= 8
6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)
= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)
= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30
= 2
7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)
= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)
= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)
= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)
= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)
TICK MÌNH NHA!
a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)
b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)
c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)
d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)
e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)
g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)
h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)
k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(\left(\sqrt{3}-\sqrt{5}\right)^2=3-2\sqrt{15}+5=8-2\sqrt{15}\)
\(\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)=7-2=5\)
\(\left(\sqrt{3}+\sqrt{5}\right)^2=3+2\sqrt{15}+5=8+2\sqrt{15}\)
By Thành Trương
\(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(B^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)
\(B^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}\)
\(B^2=8-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}\)
\(B^2=8-2\sqrt{16-7}\)
\(B^2=8-2\sqrt{9}\)
\(B^2=8-2.3\)
\(B^2=8-6\)
\(B^2=2\)
\(\Leftrightarrow B=-\sqrt{2}\)
\(C=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)\)
\(C=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{2+\sqrt{3}}\)
\(C=\left(\sqrt{3}+1\right)\sqrt{2\left(\sqrt{3}+2\right)}.\left(\sqrt{3}-2\right)\)
\(C=\left(\sqrt{3}+1\right).\sqrt{2\sqrt{3}+4}.\left(\sqrt{3}-2\right)\)
\(C=\left(\sqrt{3}+1\right).\sqrt{\left(\sqrt{3}+1\right)^2}.\left(\sqrt{3}-2\right)\)
\(C=\left(\sqrt{3}+1\right)^2.\left(\sqrt{3}-2\right)\)
\(C=4+2\sqrt{3}.\left(\sqrt{3}-2\right)\)
\(C=4\sqrt{3}-8+6-4\sqrt{3}\)
\(C=-2\)
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
\(\sqrt{\left(\sqrt{7-2}\right)^2}+\sqrt{\left(7-4\right)^2}=\sqrt{\left(\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)