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\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{2.\left(2-\sqrt{3}\right)}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{2}.\left(\sqrt{3}-1\right)}{\sqrt{2}.\sqrt{2}}=\frac{\sqrt{2}.\sqrt{3}-\sqrt{2}.1}{\sqrt{2.2}}=\frac{\sqrt{2.3}-\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{6}-\sqrt{2}}{2}\)
Sửa đề `->sqrt{7+4sqrt3}`
`=sqrt{7+4sqrt3}`
`=sqrt{4+2.2.sqrt3+3}`
`=sqrt{(2+sqrt3)^2}`
`=|2+sqrt3|`
`=2+sqrt3`
a/\(\sqrt{54}=3\sqrt{6}\)
b/\(\sqrt{50a}=\sqrt{50}.\sqrt{a}=5\sqrt{2}.\sqrt{a}\)
c/ \(\sqrt{5\left(\sqrt{3}\right)^2=}\sqrt{5.3}=\sqrt{15}\)
a) \(\sqrt{54}=\sqrt{9.6}=\sqrt{9}.\sqrt{6}=3\sqrt{6}\)
b) \(\sqrt{50a}=\sqrt{25.2a}=\sqrt{5^2.2a}=5\sqrt{2a}\)
Ta có: \(\dfrac{7-4\sqrt{3}}{\sqrt{3}-2}-\dfrac{28-10\sqrt{3}}{5-\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{3}-2\right)^2}{\sqrt{3}-2}-\dfrac{\left(5-\sqrt{3}\right)^2}{5-\sqrt{3}}\)
\(=\sqrt{3}-2-5+\sqrt{3}\)
=-7
ĐKXĐ: `x>=0;x\ne9`
`(x^2-3)/(sqrtx-3)=((x-sqrt3)(x+sqrt3))/(x+sqrt3)=x-sqrt3`
