B=\([\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}]\div[\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1]\)
a) Rút gọn
b) Tính x đế B<-1
c) Tính x để B đạt GTNN
K
Khách
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Những câu hỏi liên quan
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 \(Min_A=-3\Leftrightarrow x=0\)
BT
20 tháng 7 2017
b, tìm x thuộc Z để B thuộc Z
c, Tìm x thuộc R để B có giá trị nguyên
MN
7 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
a) \(B=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{x-9}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow B=\left(1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{x-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\frac{x-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\frac{3}{\sqrt{x}+3}:\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\frac{3}{\sqrt{x}+3}:\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(\Leftrightarrow B=\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow B=\frac{3}{\sqrt{x}+2}\)
b) ??
5 tháng 7 2019
a) \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2b}{a-b}\)
\(=\frac{a+b+2\sqrt{ab}}{2\left(a-b\right)}-\frac{a+b-2\sqrt{ab}}{2\left(a-b\right)}+\frac{4b}{2\left(a-b\right)}=\frac{a+b+2\sqrt{ab}-a-b+2\sqrt{ab}+4b}{2\left(a-b\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(a-b\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)}\)
5 tháng 7 2019
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{a-b}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)\(=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
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 \(MinB=-2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
Nhìn nhầm câu c)
\(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)làm tương tự