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a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)
\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)
\(=\sqrt{518400}\)
=720
b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)
\(=9-2\sqrt{14}+2\sqrt{14}\)
=9
c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)
\(=6+2\sqrt{5}-6\)
\(=2\sqrt{5}\)
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a) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\sqrt{15}\)
`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`
`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`
VT = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{6}\right)}{\sqrt{2}}=\sqrt{6}\) = VP (đpcm)
\(\text{a) }Ta\text{ }có:\text{ }\sqrt{5}-\sqrt{3}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{2}{\sqrt{5}+\sqrt{3}}\\ Lại\text{ }có:\text{ }\left(\sqrt{5}+\sqrt{3}\right)^2=5+3+2\sqrt{15}\\ =8+\sqrt{60}< 8+\sqrt{64}=16\\ \Rightarrow\sqrt{5}+\sqrt{3}< 4\\ \Rightarrow\dfrac{2}{\sqrt{5}+\sqrt{3}}>\dfrac{2}{4}\\ \Rightarrow\sqrt{5}-\sqrt{3}>\dfrac{1}{2}\)
\(\text{b) }\sqrt{k+1}-\sqrt{k}=\dfrac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}\\ =\dfrac{1}{\sqrt{k+1}+\sqrt{k}}\\ \Rightarrow\sqrt{7}-\sqrt{6}=\dfrac{1}{\sqrt{7}+\sqrt{6}}\\ \sqrt{6}-\sqrt{5}=\dfrac{1}{\sqrt{6}+\sqrt{5}}\\ Mà\text{ }\sqrt{7}+\sqrt{6}>\sqrt{5}+\sqrt{6}\\ \Rightarrow\dfrac{1}{\sqrt{7}+\sqrt{6}}< \dfrac{1}{\sqrt{6}+\sqrt{5}}\\\sqrt{7}-\sqrt{6}< \sqrt{6}-\sqrt{5}\)
Vậy................
6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)
7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)
8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)
9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)
