Đáp án:
Bài 4:
a.$A=\dfrac{\sqrt[]{x}}{\sqrt[]{x}+1}$
b.$A=\dfrac{3-\sqrt[]{3}}{6}$
c.$A(x-1)\ge -\dfrac{1}{4}$
Giải thích các bước giải:
Bài 4:
a.ĐKXĐ:
$\begin{cases}\sqrt[]{x}-1\ne 0 \\ \sqrt[]{x}+1\ne 0 \\x\ge 0 \\ x-1\ne 0\end{cases}\rightarrow \begin{cases}x\ge 0 \\ x\ne 1\end{cases}$
$A=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-1}-\dfrac{2}{\sqrt[]{x}+1}-\dfrac{2}{x-1}$
$\rightarrow A=\dfrac{\sqrt[]{x}.(\sqrt[]{x}+1)}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}-\dfrac{2(\sqrt[]{x}-1)}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}-\dfrac{2}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}$
$\rightarrow A=\dfrac{x+\sqrt[]{x}}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}-\dfrac{2\sqrt[]{x}-2}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}-\dfrac{2}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}$
$\rightarrow A=\dfrac{x+\sqrt[]{x}-(2\sqrt[]{x}-2)-2}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}$
$\rightarrow A=\dfrac{x-\sqrt[]{x}}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}$
$\rightarrow A=\dfrac{\sqrt[]{x}.(\sqrt[]{x}-1)}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}$
$\rightarrow A=\dfrac{\sqrt[]{x}}{\sqrt[]{x}+1}$
b.$x=7-4\sqrt[]{3}=4-4\sqrt[]{3}+3=(2-\sqrt[]{3})^2\rightarrow \sqrt[]{x}=2-\sqrt[]{3}$
$\rightarrow A=\dfrac{2-\sqrt[]{3}}{2-\sqrt[]{3}+1}=\dfrac{3-\sqrt[]{3}}{6}$
c.$A.(x-1)=A.(\sqrt[]{x}-1)(\sqrt[]{x}+1)=\dfrac{\sqrt[]{x}}{\sqrt[]{x}+1}.(\sqrt[]{x}-1)(\sqrt[]{x}+1)$
$\rightarrow A(x-1)=\sqrt[]{x}^2-\sqrt[]{x}=(\sqrt[]{x}^2-2\sqrt[]{x}.\dfrac{1}{2}+\dfrac{1}{4})-\dfrac{1}{4}\\=(\sqrt[]{x}-\dfrac{1}{2})^2-\dfrac{1}{4}\ge -\dfrac{1}{4}\quad\forall x\in đkxđ$
Dấu = xảy ra $\leftrightarrow x=\dfrac{1}{\sqrt[]{2}}$
