A=√x/√x+3 +2√x/√x-3 – 3x+9/x-9 tìm x để A=1/3 27/10/2021 Bởi Nevaeh A=√x/√x+3 +2√x/√x-3 – 3x+9/x-9 tìm x để A=1/3
Đáp án: *ĐKXĐ : `x≥0`, x$\neq$ 9 `A=frac{sqrt{x}}{sqrt{x}+3}+frac{2sqrt{x}}{sqrt{x}-3}-frac{3x+9}{x-9}` `=frac{sqrt{x}(sqrt{x}-3)}{x-9}+frac{2sqrt{x}(sqrt{x}+3)}{x-9}-frac{3x+9}{x-9}` `=frac{x-3sqrt{x}+2x+6sqrt{x}-3x-9}{x-9}` `=frac{3sqrt{x}-9}{x-9}` `A=1/3` `⇔frac{3sqrt{x}-9}{x-9}=frac{1}{3}` `⇔frac{3(sqrt{x}-3)}{(sqrt{x}-3)(sqrt{x}+3)}=frac{1}{3}` `⇔frac{3}{sqrt{x}+3}=frac{1}{3}` `⇔sqrt{x}+3=9` `⇔sqrt{x}=6` `⇔x=36(TM)` Vậy `x=36` $\text{Shield Knight}$ Bình luận
$\text{ĐKXĐ:} \ x\ge0;x\ne9$ $A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\=\dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\\=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{(\sqrt{x}-3)(\sqrt{x}+3)}\\=\dfrac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}=\dfrac{3\sqrt{x}-9}{x-9}\\\text{Ta có:} \ A=\dfrac13\\\Leftrightarrow \dfrac{3\sqrt{x}-9}{x-9}=\dfrac13\\\Leftrightarrow 9\sqrt{x}-27=x-9\\\Leftrightarrow x-9\sqrt{x}+18=0\\\Leftrightarrow (\sqrt{x}-3)(\sqrt{x}-6)=0\\\Leftrightarrow \left[\begin{array}{l}\sqrt{x}-3=0\\\sqrt{x}-6=0\end{array}\right. \Leftrightarrow \left[\begin{array}{l}\sqrt{x}=3\\\sqrt{x}=6\end{array}\right.\Leftrightarrow\left[\begin{array}{l}x=9\\x=36\end{array}\right.\\\text{Mà} \ x\ne9\Rightarrow x=36$ Vậy $x=36$ Bình luận
Đáp án:
*ĐKXĐ : `x≥0`, x$\neq$ 9
`A=frac{sqrt{x}}{sqrt{x}+3}+frac{2sqrt{x}}{sqrt{x}-3}-frac{3x+9}{x-9}`
`=frac{sqrt{x}(sqrt{x}-3)}{x-9}+frac{2sqrt{x}(sqrt{x}+3)}{x-9}-frac{3x+9}{x-9}`
`=frac{x-3sqrt{x}+2x+6sqrt{x}-3x-9}{x-9}`
`=frac{3sqrt{x}-9}{x-9}`
`A=1/3`
`⇔frac{3sqrt{x}-9}{x-9}=frac{1}{3}`
`⇔frac{3(sqrt{x}-3)}{(sqrt{x}-3)(sqrt{x}+3)}=frac{1}{3}`
`⇔frac{3}{sqrt{x}+3}=frac{1}{3}`
`⇔sqrt{x}+3=9`
`⇔sqrt{x}=6`
`⇔x=36(TM)`
Vậy `x=36`
$\text{Shield Knight}$
$\text{ĐKXĐ:} \ x\ge0;x\ne9$
$A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\=\dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\\=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{(\sqrt{x}-3)(\sqrt{x}+3)}\\=\dfrac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}=\dfrac{3\sqrt{x}-9}{x-9}\\\text{Ta có:} \ A=\dfrac13\\\Leftrightarrow \dfrac{3\sqrt{x}-9}{x-9}=\dfrac13\\\Leftrightarrow 9\sqrt{x}-27=x-9\\\Leftrightarrow x-9\sqrt{x}+18=0\\\Leftrightarrow (\sqrt{x}-3)(\sqrt{x}-6)=0\\\Leftrightarrow \left[\begin{array}{l}\sqrt{x}-3=0\\\sqrt{x}-6=0\end{array}\right. \Leftrightarrow \left[\begin{array}{l}\sqrt{x}=3\\\sqrt{x}=6\end{array}\right.\Leftrightarrow\left[\begin{array}{l}x=9\\x=36\end{array}\right.\\\text{Mà} \ x\ne9\Rightarrow x=36$
Vậy $x=36$