Toán Lớp 9: B= $\frac{1}{\sqrt{3}+2}$ – $\frac{1}{\sqrt{3}-2}$
Tính B
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Toán Lớp 9: B= $\frac{1}{\sqrt{3}+2}$ – $\frac{1}{\sqrt{3}-2}$
Tính B
Comments ( 2 )
B=(sqrt{3}-2)/((sqrt{3}+2)(sqrt{3}-2))-(sqrt{3}+2)/((sqrt{3}-2)(sqrt{3}+2))
B=(sqrt{3}-2)/((sqrt{3})^2-2^2)-(sqrt{3}+2)/((sqrt{3})^2-2^2)
B=(sqrt{3}-2)/(3-4)-(sqrt{3}+2)/(3-4)
B=(sqrt{3}-2)/(-1)-(sqrt{3}+2)/(-1)
B=(sqrt{3}-2-sqrt{3}-2)/(-1)
B=(-4)/(-1)
B=4
Vậy B=4