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a) \(A=\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}=2\sqrt{2}-2\)
b) \(B=\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)
\(=2\sqrt{2}-7+3-2\sqrt{2}=-4\)
c) \(C=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)
d) \(D=\sqrt{9+12\sqrt{2}+8}+\sqrt{9-12\sqrt{2}+8}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}=\left(3+2\sqrt{2}\right)-\left(3-2\sqrt{2}\right)=4\sqrt{2}\)
\(A=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2}{9\sqrt{3}-11\sqrt{2}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(9\sqrt{3}+11\sqrt{3}\right)\left(5-2\sqrt{6}\right)^2\)
\(=\left(49+20\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2=\left(5+2\sqrt{6}\right)^2\left(5-2\sqrt{6}\right)^2=1\)
\(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{4+5}=3\)
\(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
a: \(A=\dfrac{a\sqrt{a}+b\sqrt{b}}{a-\sqrt{ab}+b}=\sqrt{a}+\sqrt{b}=5+7=12\)
b: \(B=\left|\sqrt{c}-1\right|-\sqrt{2}=1-\sqrt{2}\)
Bài 1:
a) \(=5.|2a|-5a^2\)
b) \(=7\left(a-1\right)+5a=12a-7\)
c) \(|a-2|-5\sqrt{a+2}\)
Bài 2:
a) \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)
b) \(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c) \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
a) \(5\sqrt{4a^2}-5a^2\)
\(=5.|2a|-5a^2\)
b) \(7\sqrt{\left(a-1\right)^2}+5a\)
\(=7\left(a-1\right)+5a\)
\(=12a-7\)
c) \(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\)
\(=|a-2|-5\sqrt{a+2}\)
bài 2:
a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=3-\sqrt{2}+5-\sqrt{2}\)
\(=8-2\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
3 bài đầu dễ tự làm nhé.
Bài 4:
\(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(1-\sqrt{2}\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(1+\sqrt{2}\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{1+\sqrt{2}}{3+2\sqrt{2}}\)
\(=\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(1+\sqrt{2}\right)\left(3-2\sqrt{2}\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(3-2\sqrt{2}+3\sqrt{2}-4\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(-1+\sqrt{2}\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}+1-\sqrt{2}\)
\(=0+2\)
\(=2\)
Vậy B là số tự nhiên.
1.
a) nhân cả tử lẫn mẫu với 1+ \(\sqrt{2}-\sqrt{5}\)
b) tương tự a
2.
a) tách 29 = 20 + 9 là ra hằng đẳng thức, tiếp tục.
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
= \(2\left(\sqrt{2}-1\right)\)
b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)
Ta có:
B2 = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)
= \(12-2\sqrt{36-20}\)
= \(12-8\)
= \(4\)
\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2
Vậy B = 2