Question

[Kynz Zanz]

Cho biểu thức A=\(\dfrac{4\sqrt{x}}{x-1}\); B=\(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-1}\left(x\ge0;x\ne1\right)\)

b, Rút gọn biểu thức B

c, Tìm x để \(A.B\le8\) (giúp mình làm phần này thôi ạ, phần b không cần đâu ạ)

Answer 1

[Dang Tung]

b) \(B=\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-1}\left(x\ge0,x\ne1\right)\\ =\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}+1\right)+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ \)

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

c)

 \(AB\le8\Leftrightarrow\dfrac{4\sqrt{x}}{x-1}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\le8\\ \Leftrightarrow\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\le8\\ \Leftrightarrow\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\le8\\ \Leftrightarrow4\sqrt{x}\le8\left(x-2\sqrt{x}+1\right)\\ \) ( Nhân cả 2 vế BPT cho \(\left(\sqrt{x}-1\right)^2>0\) )

\(\Leftrightarrow8x-16\sqrt{x}+8\ge4\sqrt{x}\\ \Leftrightarrow8x-20\sqrt{x}+8\ge0\\ \Leftrightarrow2x-5\sqrt{x}+2\ge0\\ \)

\(\Leftrightarrow\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}-2\right)\ge0\\ \Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)\ge0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)\ge0\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2\ge0\\2\sqrt{x}-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2\le0\\2\sqrt{x}-1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}\ge2\\\sqrt{x}\ge\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}\le2\\\sqrt{x}\le\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4\\x\ge\dfrac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4\\x\le\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le\dfrac{1}{4}\end{matrix}\right.\)

Kết hợp ĐK: \(x\ge0,x\ne1\)

Kết luận: \(x\ge4\) hoặc \(0\le x\le\dfrac{1}{4}\) thì \(AB\le8\)

Answer 2

[Akai Haruma]

Lời giải:

ĐK: $x\geq 0; x\neq 1$

$AB=\frac{4\sqrt{x}}{(\sqrt{x}-1)^2}\leq 8$

$\Rightarrow 4\sqrt{x}\leq 8(\sqrt{x}-1)^2$
$\Leftrightarrow \sqrt{x}\leq 2(\sqrt{x}-1)^2$
$\Leftrightarrow \sqrt{x}\leq 2(x-2\sqrt{x}+1)$

$\Leftrightarrow 2x-5\sqrt{x}+2\geq 0$

$\Leftrightarrow (\sqrt{x}-2)(2\sqrt{x}-1)\geq 0$

$\Leftrightarrow \sqrt{x}\geq 2$ hoặc $\sqrt{x}\leq \frac{1}{2}$

$\Leftrightarrow x\geq 4$ hoặc $0\leq x\leq \frac{1}{4}$

Kết hợp đkxđ suy ra $x\geq 4$ hoặc $0\leq x\leq \frac{1}{4}$