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a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
= \(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(2-\sqrt{3}+\sqrt{3}-1\) = \(1\)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+2\sqrt{6}-3\) = \(\sqrt{6}\)
c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
\(a,\sqrt{3+2\sqrt{2}}\)\(+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)\(+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}\)\(+1+2-\sqrt{2}\)
\(=\left(\sqrt{2}-\sqrt{2}\right)+1+2\)
\(=0+1+2=3\)
a) \(\sqrt{3+2\sqrt{2}}\)+\(\sqrt{6-4\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}+1\right)^2}\)+\(\sqrt{\left(2-\sqrt{2}\right)^2}\)=\(\sqrt{2} +1+2-\sqrt{2}\)=3
b)\(\left(\sqrt{5}-2\sqrt{6}+\sqrt{2}\right)\sqrt{3}\)
=\((\sqrt{\left(\sqrt{3})-\sqrt{2}\right)^2}+\sqrt{2})\sqrt{3}\)
=\(\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\)
=\(\sqrt{3}.\sqrt{3}=3\)
a) = 3
b)=3
a) đáp án: 3 b) đáp án: 3
\(a.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+1+2-\sqrt{2}\)
\(=3\)
b.\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\)
\(=\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\)
\(=\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\)
\(=\sqrt{3}.\sqrt{3}\)
=3
\(\sqrt{2}+1+2-\sqrt{2=3}\)\(\sqrt{3}.\sqrt{3}=3\)
a) căn 3+2 căn 2+căn 6-4 căn 2
=căn (căn 2+1)^2+căn(2- căn 2)^2
=căn 2+1+2-căn2
=(căn 2- căn 2)+1+2
=0+1+2
=3
a) =3
b) =3
a) = 3
b) = 3
a\()\) \(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
=\(\sqrt{(\sqrt{2+1)}^2}\)+ \(\sqrt{(2-\sqrt{2)}^2}\)
=`\(\sqrt{2}\)+1 +2 -\(\sqrt{2}\)=3
b\()\)\((\sqrt{5-2\sqrt{6}}+\sqrt{2})\sqrt{3}\)
=\((\sqrt{(\sqrt{3}-\sqrt{2})^2}+\sqrt{2})\sqrt{3}\)
=\((\sqrt{3}-\sqrt{2}+\sqrt{2})\sqrt{3}\)
\(\sqrt{3}.\sqrt{3}=3\)
a)\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=3\)
b)\(\left(\sqrt{5-2\sqrt{6}+\sqrt{2}}\right)\sqrt{3}=3\)
a)\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
=\(\sqrt{2}+1+2-\sqrt{2}\)
=3
b)\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\)
=\(\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2+\sqrt{2}}\right)\sqrt{3}\)
=\(\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\)
=\(\sqrt{3}\sqrt{3}\)
=3
a, \(\sqrt{3+2\sqrt{2}}\)+ \(\sqrt{6-4\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}+1\right)^2}\)+\(\sqrt{\left(2-\sqrt{2}\right)^2}\)
=\(\sqrt{2}\)+ 1 + 2 - \(\sqrt{2}\)
=3
b, \(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\)
=\(\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\)
=\(\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\)
=\(\sqrt{3}.\sqrt{3}\)
=3
a) = 3
b) =3
3
a) 3
b) 3
a) = 3
b) = 3
a)= \(\sqrt{3}\)
b)= \(\sqrt{3}\)
a) \(\sqrt{3+2\sqrt{2}}\)+ \(\sqrt{6-4\sqrt{2}}\)
= \(\sqrt{2+1+2\sqrt{2}}\)+ \(\sqrt{4+2-2.2\sqrt{2}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}\)+ \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
=\(\sqrt{2}\)+ 1 + 2 - \(\sqrt{2}\)
= 3
a) 2\(\sqrt{2}\)+3
b) 3
a. 3
b.3
3
3-\(\sqrt{6}\)
a) 3
b) 3
a) 3
b) 3
a,= \(\sqrt{\left(\sqrt{2}+1\right)^{ }2}\)+\(\sqrt{\left(2-\sqrt{2}\right)^{ }2}\)
=\(\sqrt{2}+1+2-\sqrt{2}\)
=3
b, =\(\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^{ }}2+\sqrt{2}\right)\sqrt{3}\)
=\(\)3
a) đáp án:3
B)đáp án:3