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a) \(1+\sqrt{3}+\sqrt{5}+\sqrt{15}\)
\(=\left(1+\sqrt{3}\right)+\sqrt{5}\left(1+\sqrt{3}\right)\)
\(=\left(1+\sqrt{3}\right)\left(1+\sqrt{5}\right)\)
b) \(\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}\)
\(=\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{7}\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{7}\right)\)
c) \(\sqrt{35}-\sqrt{15}+\sqrt{14}-\sqrt{6}\)
\(=\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{2}\right)\)
e) \(xy+y\sqrt{x}+\sqrt{x}+1\)
\(=y\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)\left(y\sqrt{x}+1\right)\)
g) \(3+\sqrt{x}+9-x\)
\(=\left(3+\sqrt{x}\right)+\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)
\(=\left(3+\sqrt{x}\right)\left(4-\sqrt{x}\right)\)
Bài 3:
a) Ta có: \(4+2\sqrt{3}\)
\(=3+2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}+1\right)^2\)
b) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
c) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)
d) Ta có: \(31+10\sqrt{6}\)
\(=25+2\cdot5\cdot\sqrt{6}+6\)
\(=\left(5+\sqrt{6}\right)^2\)
e) Ta có: \(13+4\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot1+1\)
\(=\left(2\sqrt{3}+1\right)^2\)
g) Ta có: \(21+12\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot3+9\)
\(=\left(2\sqrt{3}+3\right)^2\)
h) Ta có: \(29+12\sqrt{5}\)
\(=20+2\cdot2\sqrt{5}\cdot3+3\)
\(=\left(2\sqrt{5}+3\right)^2\)
i) Ta có: \(49+8\sqrt{3}\)
\(=48+2\cdot4\sqrt{3}\cdot1\)
\(=\left(4\sqrt{3}+1\right)^2\)
k) Sửa đề: \(14-6\sqrt{5}\)
Ta có: \(14-6\sqrt{5}\)
\(=9-2\cdot3\cdot\sqrt{5}+5\)
\(=\left(3-\sqrt{5}\right)^2\)
l) Ta có: \(23-8\sqrt{7}\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=\left(4-\sqrt{7}\right)^2\)
m) Ta có: \(15-4\sqrt{11}\)
\(=11-2\cdot\sqrt{11}\cdot2+4\)
\(=\left(\sqrt{11}-2\right)^2\)
n) Sửa đề: \(28-10\sqrt{3}\)
Ta có: \(28-10\sqrt{3}\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=\left(5-\sqrt{3}\right)^2\)
o) Ta có: \(17-12\sqrt{2}\)
\(=9-2\cdot3\cdot2\sqrt{2}+8\)
\(=\left(3-2\sqrt{2}\right)^2\)
p) Ta có: \(43-30\sqrt{2}\)
\(=25-2\cdot5\cdot3\sqrt{2}+18\)
\(=\left(5-3\sqrt{2}\right)^2\)
q) Ta có: \(51-10\sqrt{2}\)
\(=50-2\cdot5\sqrt{2}\cdot1\)
\(=\left(5\sqrt{2}-1\right)^2\)
r) Ta có: \(49-12\sqrt{5}\)
\(=45-2\cdot3\sqrt{5}\cdot2+4\)
\(=\left(3\sqrt{5}-2\right)^2\)
a) \(3\sqrt{8}-4\sqrt{18}+5\sqrt{32}-\sqrt{50}=3\sqrt{4.2}-4\sqrt{9.2}+5\sqrt{16.2}-\sqrt{25.2}=6\sqrt{2}-12\sqrt{2}+20\sqrt{2}-5\sqrt{2}=9\sqrt{2}\)b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(15\sqrt{50}+5\sqrt{50.4}-3\sqrt{50.9}\right):10=\left(15\sqrt{50}+10\sqrt{50}-9\sqrt{50}\right):10=\dfrac{16\sqrt{50}}{10}=\dfrac{16\sqrt{25.2}}{10}=\dfrac{80\sqrt{2}}{10}=8\sqrt{2}\) c) \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}=2\sqrt{7.4}+2\sqrt{7.9}-3\sqrt{7.25}+\sqrt{7.16}=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}=-\sqrt{7}\)
d)
\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-2.3\sqrt{2.14}+18+6\sqrt{28}=32-6\sqrt{28}+6\sqrt{28}=32\)
1) \(5-2\sqrt{6}=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
2) \(8+2\sqrt{15}=\left(\sqrt{5}\right)^2+2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{5}+\sqrt{3}\right)^2\)
3) \(10-2\sqrt{21}=\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{7}-\sqrt{3}\right)^2\)
4) \(21+6\sqrt{6}=\left(\sqrt{18}\right)^2+2.\sqrt{18}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{18}+\sqrt{3}\right)^2\)
5) \(14+8\sqrt{3}=\left(\sqrt{8}\right)^2+2.\sqrt{8}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2\)
6) \(36-12\sqrt{5}=\left(\sqrt{30}\right)^2-2.\sqrt{30}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{30}-\sqrt{6}\right)^2\)
7) \(25+4\sqrt{6}=\left(\sqrt{24}\right)^2+2\sqrt{24}.1+1^2=\left(\sqrt{24}+1\right)^2\)
8) \(98-16\sqrt{3}=\left(\sqrt{96}\right)^2-2\sqrt{96}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{96}-\sqrt{2}\right)^2\)
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
a) \(21-8\sqrt{5}=16-2\times4\times\sqrt{5}+5=\left(4-\sqrt{5}\right)^2\)
b) \(47-12\sqrt{11}=36-2\times6\times\sqrt{11}+11=\left(6-\sqrt{11}\right)^2\)
c) \(13-4\sqrt{3}=12-2\times1\times\sqrt{3}+1=\left(2\sqrt{3}-1\right)^2\)
d) \(43+30\sqrt{2}=25+2\times5\times3\sqrt{2}+18=\left(5+3\sqrt{2}\right)^2\)
e) \(41+24\sqrt{2}=9+2\times3\times4\sqrt{2}+32=\left(3+4\sqrt{2}\right)^2\)
g) \(29-12\sqrt{5}=9+2\times3\times2\sqrt{5}+20=\left(3+2\sqrt{5}\right)^2\)
h) \(49-8\sqrt{3}=48-2\times4\sqrt{3}\times1+1=\left(4\sqrt{3}-1\right)^2\)
i) \(37-12\sqrt{7}=28-2\times3\times2\sqrt{7}+9=\left(2\sqrt{7}-3\right)^2\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
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