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22 tháng 6 2019

Căn bậc hai. Căn bậc ba

22 tháng 6 2019

a) \(19+8\sqrt{3}=3+2\sqrt{3}\cdot4+16=\left(\sqrt{3}+4\right)^2\)

b) \(11-4\sqrt{6}=3-2\sqrt{3}\cdot2\sqrt{2}+8=\left(\sqrt{3}-2\sqrt{2}\right)^2\)

c) \(9-4\sqrt{2}=8-2\cdot2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)

d) \(21+6\sqrt{10}=18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5-2=\left(3\sqrt{2}+\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2\)

e) \(23+6\sqrt{10}=18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5=\left(3\sqrt{2}+\sqrt{5}\right)^2\)

f) \(49-20\sqrt{6}=\left(5\sqrt{2}\right)^2-2\cdot5\sqrt{2}\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-13=\left(5\sqrt{2}-2\sqrt{3}\right)^2-\left(\sqrt{13}\right)^2\)

14 tháng 5 2018

1/ \(7-2\sqrt{6}=\left(\sqrt{6}\right)^2-2\sqrt{6}+1\)

\(=\left(\sqrt{6}-1\right)^2\)

2/ \(10+2\sqrt{21}=\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2\)

\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)

4/ \(10+4\sqrt{6}=2^2+2.2.\sqrt{6}+\left(\sqrt{6}\right)^2\)

\(=\left(2+\sqrt{6}\right)^2\)

5/ \(11-2\sqrt{30}=\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2\)

= \(\left(\sqrt{6}-\sqrt{5}\right)^2\)

8/ \(11+4\sqrt{7}=2^2+2.2.\sqrt{7}+\left(\sqrt{7}\right)^2\)

= \(\left(2+\sqrt{7}\right)^2\)

10/ \(12+6\sqrt{3}=3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)

= \(\left(3+\sqrt{3}\right)^2\)

15 tháng 6 2016

A = ((20 + 1) . 20 : 2) . 2 = 420

B = (25 + 20) . 6  : 2 = 135

C = ( 33 + 26) . 8 : 2 = 236

D = (1 + 100) .100 : 2 = 5050

15 tháng 6 2016

Toán lướp 9 dễ như vậy à bạn

14 tháng 6 2018

Các câu sau bạn tự làm đi mCăn thức bậc hai và hằng đẳng thức căn bậc hai của bình phương

7 tháng 7 2018

a, \(\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}\right)^2-2\sqrt{3}.\sqrt{5}-\left(\sqrt{3}\right)^2\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\sqrt{5}-\sqrt{3}\)

b,

16 tháng 6 2018

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) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)

\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)

\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)

\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)

\(=5\sqrt{7}\)

16 tháng 6 2018

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

NV
17 tháng 6 2019

\(A=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2}{9\sqrt{3}-11\sqrt{2}}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)\left(9\sqrt{3}+11\sqrt{3}\right)\left(5-2\sqrt{6}\right)^2\)

\(=\left(49+20\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2=\left(5+2\sqrt{6}\right)^2\left(5-2\sqrt{6}\right)^2=1\)

\(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{4+5}=3\)

\(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)