Question

SOS

Answer 1

`A=\sqrt{27}-\sqrt{12}`

`A=\sqrt{3^2 .3}-\sqrt{2^2 .3}`

`A=3\sqrt{3}-2\sqrt{3}=\sqrt{3}`

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`B=\sqrt{a}-[a-4]/[\sqrt{a}+2]`             `a >= 0`

`B=\sqrt{a}-[(\sqrt{a}-2)(\sqrt{a}+2)]/[\sqrt{a}+2]`

`B=\sqrt{a}-\sqrt{a}+2=2`

Answer 2

\(a,A=\sqrt{27}-\sqrt{12}\)

\(A=\sqrt{3^3}-\sqrt{2^2.3}=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\)

\(b,B=\sqrt{a}-\dfrac{a-4}{\sqrt{a}+2}\) với \(a\ge0\)

\(B=\sqrt{a}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}+2}\)

\(B=\sqrt{a}-\left(\sqrt{a}-2\right)\)

\(B=\sqrt{a}-\sqrt{a}+2\)

\(B=2\)