Question

Cộng các đa thức sau:

Answer 1

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne-1\\x\ne0\end{matrix}\right.\Leftrightarrow}x>0}\)

b,\(\left(\dfrac{1\left(\sqrt{x}\right)-1\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}\right):\left(\dfrac{1\left(\sqrt{x}+1\right)-1\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

=\(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)=\left(\dfrac{1}{\left(\sqrt{x}-1\right)}\right):\left(\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)=\dfrac{1}{\left(\sqrt{x}-1\right)\sqrt{x}}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

=\(\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

c, Thay x=4 vào biểu thức P ta có : \(\dfrac{\sqrt{4}+1}{2\sqrt{4}}=\dfrac{2+1}{\sqrt{16}}=\dfrac{3}{4}\)

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