B=((x+3)/(x-9)+(1)/(\sqrt(x)+3))-:(\sqrt(x))/(\sqrt(x)-3) a.Rút gọn B b.Tính B khi x=3+2\sqrt2 c.c/m B>1/3 02/07/2021 Bởi Peyton B=((x+3)/(x-9)+(1)/(\sqrt(x)+3))-:(\sqrt(x))/(\sqrt(x)-3) a.Rút gọn B b.Tính B khi x=3+2\sqrt2 c.c/m B>1/3
Đáp án: $B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$ Giải thích các bước giải: a) $B=\bigg(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\bigg):\dfrac{\sqrt{x}}{\sqrt{x}-3}$ ĐK: $x\neq9;\;x>0$ $B=\bigg(\dfrac{x+3}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{1}{\sqrt{x}+3}\bigg)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}$ $B=\dfrac{x+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}$ $B=\dfrac{x+\sqrt{x}}{\sqrt{x}(\sqrt{x}+3)}$ $B=\dfrac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+3)}$ $B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$ Vậy $B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$ với $x\neq9;\;x>0$ b) Thay $x=3+2\sqrt{2}$ (TM) vào biểu thức B ta được: $B=\dfrac{\sqrt{3+2\sqrt{2}}+1}{\sqrt{3+2\sqrt{2}}+3}$ $=\dfrac{\sqrt{2+2\sqrt{2}+1}+1}{\sqrt{2+2\sqrt{2}+1}+3}$ $=\dfrac{\sqrt{(\sqrt{2}+1)^2}+1}{\sqrt{(\sqrt{2}+1)^2}+3}$ $=\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1+3}$ $=\dfrac{2+\sqrt{2}}{4+\sqrt{2}}$ $=\dfrac{(2+\sqrt{2})(4-\sqrt{2})}{(4+\sqrt{2})(4-\sqrt{2})}$ $=\dfrac{8-2\sqrt{2}+4\sqrt{2}-2}{16-2}$ $=\dfrac{6+2\sqrt{2}}{14}$ $=\dfrac{3+\sqrt{2}}{7}$ Vậy $B=\dfrac{3+\sqrt{2}}{7}$ với $x=3+2\sqrt{2}$ c) Xét hiệu $B-\dfrac{1}{3}$ ($x\neq9;\;x>0$) $=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{3}$ $=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{3(\sqrt{x}+3)}$ $=\dfrac{2\sqrt{x}}{3\sqrt{x}+9}$ Mà $x>0\to\sqrt{x}>0\to\begin{cases}2\sqrt{x}>0\\3\sqrt{x}+9>0\end{cases}\to B-\dfrac{1}{3}>0\to B>\dfrac{1}{3}$ (đpcm) Bình luận
Đáp án: c. \(B > \dfrac{1}{3}\) Giải thích các bước giải: \(\begin{array}{l}a.DK:x > 0;x \ne 9\\B = \left( {\dfrac{{x + 3}}{{x – 9}} + \dfrac{1}{{\sqrt x + 3}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x – 3}}} \right)\\ = \left[ {\dfrac{{x + 3 + \sqrt x – 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right].\dfrac{{\sqrt x – 3}}{{\sqrt x }}\\ = \dfrac{{x + \sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x }}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\b.Thay:x = 3 + 2\sqrt 2 = 2 + 2\sqrt 2 .1 + 1\\ = {\left( {\sqrt 2 + 1} \right)^2}\\ \to B = \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 1}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 3}} = \dfrac{{\sqrt 2 + 1 + 1}}{{\sqrt 2 + 1 + 3}}\\ = \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 + 4}} = \dfrac{{3 + \sqrt 2 }}{7}\\c.Xét:B – \dfrac{1}{3} = \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}} – \dfrac{1}{3}\\ = \dfrac{{3\sqrt x + 3 – \sqrt x – 3}}{{3\left( {\sqrt x + 3} \right)}} = \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}}\\Do:x > 0\\ \to \left\{ \begin{array}{l}2\sqrt x > 0\\\sqrt x + 3 > 0\end{array} \right.\\ \to \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} > 0\\ \to B > \dfrac{1}{3}\end{array}\) Bình luận
Đáp án:
$B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$
Giải thích các bước giải:
a) $B=\bigg(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\bigg):\dfrac{\sqrt{x}}{\sqrt{x}-3}$
ĐK: $x\neq9;\;x>0$
$B=\bigg(\dfrac{x+3}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{1}{\sqrt{x}+3}\bigg)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}$
$B=\dfrac{x+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}$
$B=\dfrac{x+\sqrt{x}}{\sqrt{x}(\sqrt{x}+3)}$
$B=\dfrac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+3)}$
$B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$
Vậy $B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$ với $x\neq9;\;x>0$
b) Thay $x=3+2\sqrt{2}$ (TM) vào biểu thức B ta được:
$B=\dfrac{\sqrt{3+2\sqrt{2}}+1}{\sqrt{3+2\sqrt{2}}+3}$
$=\dfrac{\sqrt{2+2\sqrt{2}+1}+1}{\sqrt{2+2\sqrt{2}+1}+3}$
$=\dfrac{\sqrt{(\sqrt{2}+1)^2}+1}{\sqrt{(\sqrt{2}+1)^2}+3}$
$=\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1+3}$ $=\dfrac{2+\sqrt{2}}{4+\sqrt{2}}$
$=\dfrac{(2+\sqrt{2})(4-\sqrt{2})}{(4+\sqrt{2})(4-\sqrt{2})}$
$=\dfrac{8-2\sqrt{2}+4\sqrt{2}-2}{16-2}$ $=\dfrac{6+2\sqrt{2}}{14}$
$=\dfrac{3+\sqrt{2}}{7}$
Vậy $B=\dfrac{3+\sqrt{2}}{7}$ với $x=3+2\sqrt{2}$
c) Xét hiệu $B-\dfrac{1}{3}$ ($x\neq9;\;x>0$)
$=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{3}$
$=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{3(\sqrt{x}+3)}$
$=\dfrac{2\sqrt{x}}{3\sqrt{x}+9}$
Mà $x>0\to\sqrt{x}>0\to\begin{cases}2\sqrt{x}>0\\3\sqrt{x}+9>0\end{cases}\to B-\dfrac{1}{3}>0\to B>\dfrac{1}{3}$ (đpcm)
Đáp án:
c. \(B > \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 9\\
B = \left( {\dfrac{{x + 3}}{{x – 9}} + \dfrac{1}{{\sqrt x + 3}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x – 3}}} \right)\\
= \left[ {\dfrac{{x + 3 + \sqrt x – 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right].\dfrac{{\sqrt x – 3}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x }}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
b.Thay:x = 3 + 2\sqrt 2 = 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to B = \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 1}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 3}} = \dfrac{{\sqrt 2 + 1 + 1}}{{\sqrt 2 + 1 + 3}}\\
= \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 + 4}} = \dfrac{{3 + \sqrt 2 }}{7}\\
c.Xét:B – \dfrac{1}{3} = \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}} – \dfrac{1}{3}\\
= \dfrac{{3\sqrt x + 3 – \sqrt x – 3}}{{3\left( {\sqrt x + 3} \right)}} = \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}}\\
Do:x > 0\\
\to \left\{ \begin{array}{l}
2\sqrt x > 0\\
\sqrt x + 3 > 0
\end{array} \right.\\
\to \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} > 0\\
\to B > \dfrac{1}{3}
\end{array}\)