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√20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)
\(=\left(7-6+1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)
\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
\(=\left(3-4+7\right)\sqrt{a}\)
\(=6\sqrt{a}\)
4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)
\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
\(=4\sqrt{b}-5\sqrt{10b}\)
Bài 1:
a: \(\sqrt{50}+2\sqrt{8}-\dfrac{3}{2}\cdot\sqrt{72}+\sqrt{125}\)
\(=5\sqrt{2}+2\cdot2\sqrt{2}-\dfrac{3}{2}\cdot6\sqrt{2}+\sqrt{125}\)
\(=9\sqrt{2}-9\sqrt{2}+5\sqrt{5}=5\sqrt{5}\)
b: \(\left(3\sqrt{2}-\sqrt{5}\right)^2-\dfrac{9}{\sqrt{5}-\sqrt{2}}\)
\(=18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5-\dfrac{9\left(\sqrt{5}+\sqrt{2}\right)}{5-2}\)
\(=23-6\sqrt{10}-3\left(\sqrt{5}+\sqrt{2}\right)\)
\(=23-6\sqrt{10}-3\sqrt{5}-3\sqrt{2}\)
c: \(5\sqrt{4a}-3\sqrt{25a}+\sqrt{9a}\)
\(=5\cdot2\sqrt{a}-3\cdot5\sqrt{a}+3\sqrt{a}\)
\(=10\sqrt{a}-15\sqrt{a}+3\sqrt{a}=-2\sqrt{a}\)
Vì đây toàn là số cụ thể rồi nên không có đkxđ bạn nhé.
Lời giải:
a.
$=\sqrt{2}+4\sqrt{2}+6\sqrt{2}-3\sqrt{2}=8\sqrt{2}$
b.
$=\frac{13(5-2\sqrt{3})}{(5+2\sqrt{3})(5-2\sqrt{3})}+2\sqrt{3}=\frac{13(5-2\sqrt{3})}{13}+2\sqrt{3}$
$=5-2\sqrt{3}+2\sqrt{3}=5$
c.
$=2\sqrt{5}-|2-\sqrt{5}|=2\sqrt{5}-(\sqrt{5}-2)=\sqrt{5}+2$
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
(vì a < 0 nên |a| = -a, b2 > 0 với mọi b ≠ 0 nên |b2| = b2 )
b: \(x-2\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2\)
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
