Bài 1:
Cho A= ( 2x+1/(x√x)-1 – 1/ (√x)-1) : (1 – x-2/x+√x+1)
a,Rút gọn A
b, Tìm x biết x= 2-√3/2
c, Tìm x ∈ Z để A ∈ Z
d, Tìm giá trị nhỏ nhất của A
e, Tìm x để A= 1/3
f, So sánh A với 1
g, Tìm x để A lớn hơn 1/2
Mình viết dấu ngoặc cho x√x và √x cho dễ nhìn thôi chứ ko có ngoặc đâu nha
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{2x + 1}}{{x\sqrt x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\left( {1 – \dfrac{{x – 2}}{{x + \sqrt x + 1}}} \right)\\
= \dfrac{{2x + 1 – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 – x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{2x + 1 – x – \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{x – \sqrt x }}{{\sqrt x – 1}}.\dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
b)x = \dfrac{{2 – \sqrt 3 }}{2}\left( {tmdk} \right)\\
\Rightarrow x = \dfrac{{4 – 2\sqrt 3 }}{4} = {\left( {\dfrac{{\sqrt 3 – 1}}{2}} \right)^2}\\
\Rightarrow \sqrt x = \dfrac{{\sqrt 3 – 1}}{2}\\
\Rightarrow A = \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
= \dfrac{{\dfrac{{\sqrt 3 – 1}}{2}}}{{\dfrac{{\sqrt 3 – 1}}{2} + 3}} = \dfrac{{\sqrt 3 – 1}}{{\sqrt 3 + 5}}\\
= \dfrac{{ – 4 + 3\sqrt 3 }}{{11}}\\
c)A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 – 3}}{{\sqrt x + 3}}\\
= 1 – \dfrac{3}{{\sqrt x + 3}}\\
A \in Z\\
\Rightarrow \dfrac{3}{{\sqrt x + 3}} \in Z\\
\Rightarrow \sqrt x + 3 = 3\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Rightarrow \sqrt x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)\\
\text{Vậy}\,x = 0\\
d)A = 1 – \dfrac{3}{{\sqrt x + 3}}\\
DO:\sqrt x + 3 \ge 3\\
\Rightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Rightarrow \dfrac{3}{{\sqrt x + 3}} \le 1\\
\Rightarrow – \dfrac{3}{{\sqrt x + 3}} \ge – 1\\
\Rightarrow 1 – \dfrac{3}{{\sqrt x + 3}} \ge 0\\
\Rightarrow A \ge 0\\
\Rightarrow GTNN:A = 0\\
Khi:x = 0\\
e)A = \dfrac{1}{3}\\
\Rightarrow \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{3}\\
\Rightarrow \sqrt x + 3 = 3\sqrt x \\
\Rightarrow 2\sqrt x = 3\\
\Rightarrow \sqrt x = \dfrac{3}{2}\\
\Rightarrow x = \dfrac{9}{4}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{9}{4}\\
f)A – 1 = \dfrac{{\sqrt x }}{{\sqrt x + 3}} – 1\\
= \dfrac{{\sqrt x – \sqrt x – 3}}{{\sqrt x + 3}} = \dfrac{{ – 3}}{{\sqrt x + 3}} < 0\\
\Rightarrow A – 1 < 0\\
\Rightarrow A < 1
\end{array}$
$\begin{array}{l}
g)A > \dfrac{1}{2}\\
\Rightarrow \dfrac{{\sqrt x }}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\Rightarrow \dfrac{{2\sqrt x – \sqrt x – 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\Rightarrow \dfrac{{\sqrt x – 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\Rightarrow \sqrt x – 3 > 0\\
\Rightarrow \sqrt x > 3\\
\Rightarrow x > 9\\
\text{Vậy}\,x > 9
\end{array}$