Câu hỏi của AF.Khánh Phương

Question

Cho biểu thức :

$A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)$

a) Rút gọn A ?

b) Tìm giá trị của A khi x = \(\df...

Đức Minh · Accepted Answer

ĐKXĐ : $x\ne1;x\ne0;x>0$ a) $A=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\cdot\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{x-1}{\sqrt{x}}\right)$ $=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4x\sqrt{x}-4\sqrt{x}}{\sqrt{x}}$ $=\dfrac{4x\sqrt{x}}{\sqrt{x}}=4x$ b) $x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}$ Suy ra : $A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}$ c) $\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x$ $\Leftrightarrow2\sqrt{x}>4x$ $\Leftrightarrow2\sqrt{x}-4x>0$ $\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0$ $\Rightarrow1-2\sqrt{x}>0$ (do x > 0 nên $2\sqrt{x}>0$) $\Leftrightarrow1>2\sqrt{x}$ $\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}$ Theo điều kiện suy ra giá trị của x để $\sqrt{A}>A$ là $0< x< \dfrac{1}{4}$Câu trả lời: ĐKXĐ : $x\ne1;x\ne0;x>0$ a) $A=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\cdot\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{x-1}{\sqrt{x}}\right)$ $=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4x\sqrt{x}-4\sqrt{x}}{\sqrt{x}}$ $=\dfrac{4x\sqrt{x}}{\sqrt{x}}=4x$ b) $x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}$ Suy ra : $A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}$ c) $\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x$ $\Leftrightarrow2\sqrt{x}>4x$ $\Leftrightarrow2\sqrt{x}-4x>0$ $\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0$ $\Rightarrow1-2\sqrt{x}>0$ (do x > 0 nên $2\sqrt{x}>0$) $\Leftrightarrow1>2\sqrt{x}$ $\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}$ Theo điều kiện suy ra giá trị của x để $\sqrt{A}>A$ là $0< x< \dfrac{1}{4}$

qwerty · Answer

a) $A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)$

$=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{x-1}{\sqrt{x}}$

$=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}$

$=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}$

$=\dfrac{4\sqrt{x}+4\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}$

$=4\sqrt{x}\cdot\left[1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]\cdot\dfrac{1}{\sqrt{x}}$

$=4\left(1+x-1\right)$

$=4x$

b) Thay $x=\dfrac{\sqrt{6}}{2+\sqrt{6}}$ vào biểu thức A.

Ta có:

$4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\
=-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12$

Vậy giá trị biểu thức A tại $x=\dfrac{\sqrt{6}}{2+\sqrt{6}}$ là $-4\sqrt{6}+12$

c) Để $\sqrt{A}>A$

$\Rightarrow\sqrt{4x}>4x$

$\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)$Câu trả lời: a) $A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)$