Rút gọn C=\(\frac{1}{3-2\sqrt{x}}\)
tim GTNN của c phay voi c =\(\frac{1}{C}.\frac{1}{\sqrt{x}+1}\)
Tinh C voi x=\(\frac{2}{2-\sqrt{3}}\)
K
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KN
20 tháng 9 2020
a) \(ĐK:x\ge0,x\ne9\)
Với\(x\ge0,x\ne9\)thì \(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right]\)\(=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)\(=\left[\frac{2x-6\sqrt{x}}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3\sqrt{x}+9}{x-9}\right]:\left[\frac{\sqrt{x}+1}{\sqrt{x}-3}\right]\)\(=\left[\frac{3x-6\sqrt{x}-9}{x-9}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)
b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
Mà \(\sqrt{x}+3>0\)nên \(4\sqrt{x}-6< 0\Leftrightarrow\sqrt{x}< \frac{3}{2}\Leftrightarrow x< \frac{9}{4}\)
Vậy với \(0\le x< \frac{9}{4}\)thì B < -1
c) \(B=\frac{4\sqrt{x}-6}{\sqrt{x}+3}=\frac{4\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=4-\frac{18}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\frac{18}{\sqrt{x}+3}\le6\Leftrightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Leftrightarrow4-\frac{18}{\sqrt{x}+3}\ge-2\)
Vậy \(MinB=-2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
20 tháng 9 2020
Nhìn nhầm câu c)
\(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)làm tương tự
KT
14 tháng 7 2018
a) ĐKXĐ: \(x\ge0;x\ne9\)
mk chỉnh lại đề bài nhé, chắc có lẽ bn ghi nhầm:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)
20 tháng 9 2020
a) đk: \(x\ge0;x\ne9\)
Ta có:
\(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]\div\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\sqrt{x}-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)
b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\) , mà \(\sqrt{x}+3\ge3>0\left(\forall x\right)\)
=> \(4\sqrt{x}-6< 0\)
\(\Leftrightarrow4\sqrt{x}< 6\)
\(\Rightarrow\sqrt{x}< \frac{3}{2}\)
\(\Rightarrow x< \frac{9}{4}\)
Vậy \(0\le x< \frac{9}{4}\)
20 tháng 9 2020
c) Ta có: \(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=3-\frac{18}{\sqrt{x}+3}\)
Vì \(\sqrt{x}+3\ge3\Rightarrow\frac{18}{\sqrt{x}+3}\le6\)
\(\Leftrightarrow3-\frac{18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow A\ge-3\)
Dấu "=" xảy ra khi: \(\sqrt{x}+3=3\Rightarrow x=0\)
Vậy \(Min_A=-3\Leftrightarrow x=0\)
15 tháng 7 2019
\(C=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x-\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\) (tự tìm ĐKXĐ)
\(=\frac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}-1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}+1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+3\)
GTNN:\(x-\sqrt{x}+3=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
\(\Rightarrow Min\left(C\right)=\frac{11}{4}khi..\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
PT
25 tháng 10 2016
\(C=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{9-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{1}{2}\right)\) ĐK \(x\ge0;x\ne9\)
\(C=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)}-\frac{1\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)}\right)\)
\(C=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\right)\)
\(C=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{2\left(\sqrt{x}-3\right)}\)
\(C=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) x \(\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(C=\frac{-6}{\sqrt{x}+3}\)
b: ta có \(C=\frac{-6}{\sqrt{x}+3}\) mà \(C=\frac{1}{2}\)
\(\frac{-6}{\sqrt{x}+3}=\frac{1}{2}\)
\(-12=\sqrt{x}+3\)
\(\sqrt{x}=-15\)(Loại)
=> x không có giá trị nào để C=\(\frac{1}{2}\)