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3:
a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
2:
a: 2căn 7=căn 28
3căn 2=căn 18
mà 28>18
nên 2*căn 7>3*căn 2
b: 5=2+3
mà 3>căn 2
nên 2+3>2+căn 2
=>5>2+căn 2
1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)
\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)
\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)
2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)
\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)
b) \(5=2+3=2+\sqrt{9}\)
Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)
3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
a) \(=5\left|a\right|+3a=5a+3a=8a\)
b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a) \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
b) \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}=2\sqrt{3}-1\)
c) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
d) \(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+2-\sqrt{2}}\)
\(=\sqrt{5+\sqrt{2}}\)
e) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)
\(=2+\sqrt{9-4\sqrt{5}}\)
\(=2+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2+\sqrt{5}-2=\sqrt{5}\)
f) đề sai nhé:
\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)
g) \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)
h) \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)
