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Câu hỏi của Toán Chuyên Học - Toán lớp 9 | Học trực tuyến
ban co the nao giai chi tiet cho minh dc ko
câu b trc nha
B = \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{2}.\sqrt{3}+2\sqrt{2}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{2+2+\sqrt{2}+2\sqrt{2}-\sqrt{3}-\sqrt{6}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)+2\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}+1\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\left(\sqrt{2}+1\right)\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\sqrt{2}\) + 1
A = \(\dfrac{21}{2}\) . (\(\sqrt{4+2\sqrt{3}}\) + \(\sqrt{6-2\sqrt{5}}\) )2 - 15\(\sqrt{15}\)
- 3(\(\sqrt{4-2\sqrt{3}}\) +\(\sqrt{6+2\sqrt{5}}\) )2
= \(\dfrac{21}{2}\).(\(\sqrt{\left(\sqrt{3}+1\right)^2}\) + \(\sqrt{\left(\sqrt{5}-1\right)^2}\))2-15\(\sqrt{15}\)
-3(\(\sqrt{\left(\sqrt{3}-1\right)^2}\) + \(\sqrt{\left(\sqrt{5}+1\right)^2}\))2
= \(\dfrac{21}{2}\).(\(\sqrt{3}\) +1+ \(\sqrt{5}\) - 1)2 -3.(\(\sqrt{3}\) - 1 + \(\sqrt{5}\) +1)2
- 15\(\sqrt{15}\)
= \(\dfrac{21}{2}\).(8+2\(\sqrt{15}\) ) - 3(8 + 2\(\sqrt{15}\) ) -15\(\sqrt{15}\)
= \(\dfrac{15}{2}\) .2.(4+\(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.( 4 + \(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.(4+\(\sqrt{15}\) -\(\sqrt{15}\)) =15.4 = 60
Vậy A = 60.
a) \(\frac{2}{3}\sqrt{3}-\frac{1}{4}\sqrt{18}+\frac{2}{5}\sqrt{2}-\frac{1}{4}\sqrt{12}\)
\(=\frac{2}{3}\sqrt{3}-\frac{1}{4}\sqrt{2\times3^2}+\frac{2}{5}\sqrt{2}-\frac{1}{4}\sqrt{3\times2^2}\)
\(=\frac{2}{3}\sqrt{3}-\frac{3}{4}\sqrt{2}+\frac{2}{5}\sqrt{2}-\frac{1}{2}\sqrt{3}\)
\(=\frac{1}{6}\sqrt{3}-\frac{7}{20}\sqrt{2}=\frac{10\sqrt{3}-21\sqrt{2}}{60}\)
b) \(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(=\sqrt{2}\left(\sqrt{5}+1\right)\left(5-2\sqrt{5}\times1+1\right)\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}\)
\(=4\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}\times1+1}\)
\(=4\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}+1\right)^2}=4\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=4^2=16\)
\(H=2\sqrt{27}+\sqrt{243}-6\sqrt{12}\\ =2\cdot\sqrt{9}\cdot\sqrt{3}+\sqrt{81}\cdot\sqrt{3}-6\cdot\sqrt{4}\cdot\sqrt{3}\\ =2\cdot3\cdot\sqrt{3}+9\cdot\sqrt{3}-6\cdot2\cdot\sqrt{3}\\ =6\sqrt{3}+9\sqrt{3}-12\sqrt{3}\\ =3\sqrt{3}=\sqrt{9}\cdot\sqrt{3}=\sqrt{27}\)
\(I=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\\ =\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\\ =\sqrt{\sqrt{13}^2-2\cdot\sqrt{13}\cdot1+1^2}+\sqrt{\sqrt{13}^2+2\cdot\sqrt{13}\cdot1+1^2}\\ =\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\\ =\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\\ =\sqrt{13}-1+\sqrt{13}+1\\ =2\sqrt{13}=\sqrt{4}\cdot\sqrt{13}=\sqrt{52}\)
\(I=\sqrt{10-4\sqrt{6}}+\sqrt{10+4\sqrt{6}}\\ =\sqrt{6-2\cdot\sqrt{6}\cdot2+4}+\sqrt{6+2\cdot\sqrt{6}\cdot2+4}\\ =\sqrt{\sqrt{6}^2-2\cdot\sqrt{6}\cdot2+2^2}+\sqrt{\sqrt{6}^2+2\cdot\sqrt{6}\cdot2+2^2}\\ =\sqrt{\left(\sqrt{6}-2\right)^2}+\sqrt{\left(\sqrt{6}+2\right)^2}\\ =\left|\sqrt{6}-2\right|+\left|\sqrt{6}+2\right|\\ =\sqrt{6}-2+\sqrt{6}+2\\ =2\sqrt{6}=\sqrt{4}\cdot\sqrt{6}=\sqrt{24}\)
B1: a) \(\sqrt{12,1.490}=\sqrt{12,1.10.49}=\sqrt{121}.\sqrt{49}=11.7=77\)
b) \(\sqrt{72.32}=\sqrt{36.2.32}=\sqrt{36}.\sqrt{64}=6.8=48\)
B2: a) \(\sqrt{48.75a^2}=\sqrt{3600a^2}=60\left|a\right|\)
b) \(\sqrt{8a^2}.\sqrt{18a^4}=\sqrt{8a^2.18a^4}=\sqrt{144a^6}=-12a^3\)
c) \(\sqrt{a}.\sqrt{\dfrac{9}{a}}=\sqrt{a.\dfrac{9}{a}}=\sqrt{9}=3\)
d) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
\(taco:A^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}}\Rightarrow A^2-A=2\Rightarrow A=2\left(dpcm\right)\)
\(A=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}}}\)
\(\Rightarrow A^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}}\)
\(\Rightarrow A^2-A=\left(2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}\right)-\left(\sqrt{2+\sqrt{2+\cdot\cdot\cdot\cdot}}\right)\)
\(\Rightarrow A=2\)