a, Với x = 4
\(A=\left(\dfrac{2+2}{2+1}-\dfrac{2.2-2}{2-1}\right)\left(4-1\right)=\left(\dfrac{4}{3}-2\right).3=-\dfrac{2}{3}.3=-2\)
b, Với x >= 0 ; x khác 1
\(A=\left(\dfrac{x+\sqrt{x}-2-2\left(x-1\right)}{x-1}\right)\left(x-1\right)=-x+\sqrt{x}\)
Ta có \(-\left(x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu ''='' xảy ra khi x = 1/4
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