Câu hỏi của WonMaengGun

Question

a) Rút gọn biểu thức:$\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}$b) Tìm giá trị nhỏ nhất của biểu thức B=$x^2-x\sqrt{3}+1$

HT.Phong (9A5) · Accepted Answer

a) $\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}$$=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)$$=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)$$=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)$$=-\left(2-5\right)$$=-\left(-3\right)$$=3$b) Ta có:$x^2-x\sqrt{3}+1$ $=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}$$=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}$Mà: $\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x$ nên$\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x$Dấu "=" xảy ra:$\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}$$\Leftrightarrow x=\dfrac{\sqrt{3}}{2}$Vậy: GTNN của biểu thức là $\dfrac{1}{4}$ tại $x=\dfrac{\sqrt{3}}{2}$Câu trả lời: a) $\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}$$=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)$$=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)$$=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)$$=-\left(2-5\right)$$=-\left(-3\right)$$=3$b) Ta có:$x^2-x\sqrt{3}+1$ $=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}$$=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}$Mà: $\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x$ nên$\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x$Dấu "=" xảy ra:$\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}$$\Leftrightarrow x=\dfrac{\sqrt{3}}{2}$Vậy: GTNN của biểu thức là $\dfrac{1}{4}$ tại $x=\dfrac{\sqrt{3}}{2}$

HaNa · Answer

a)$\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\
=\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\
=\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\
=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\
=-\left(\sqrt{2}^2-\sqrt{5}^2\right)\
=-\left(2-5\right)\
=-\left(-3\right)\
=3$Câu trả lời: a)$\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\
=\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\
=\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\
=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\
=-\left(\sqrt{2}^2-\sqrt{5}^2\right)\
=-\left(2-5\right)\
=-\left(-3\right)\
=3$

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