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HP
31 tháng 8 2021
a, ĐK: \(x\ge0;x\ne9\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=-\dfrac{3}{\sqrt{x}-3}\)
HP
31 tháng 8 2021
b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)
\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)
\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)
31 tháng 10 2021
a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)
MH
17 tháng 10 2023
\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)^2}\)
\(P=-\dfrac{1}{3}\)
\(\Rightarrow\left(\sqrt{x}+3\right)^2=3\sqrt{x}+3\)
\(\Leftrightarrow x-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x=9\left(Vì\sqrt{x}+2>0\right)\)
\(P=-\left(\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}+3\right)^2}\right)=-\left(\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)^2}\right)< -3< -1\)
YT
23 tháng 7 2018
\(P=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\) dk \(x\ge0;x\ne9\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\frac{3\sqrt{x}-9}{x-9}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)
b)
\(P=\frac{1}{3}\Leftrightarrow\frac{3}{\sqrt{x}+3}=\frac{1}{3}\Leftrightarrow\sqrt{x}+3=9\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)
vay ......................................
nếu có sai bn thông cảm nha
NM
1
21 tháng 9 2015
C = \(=\frac{x+3+2\text{ }\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x+3\right)\left(x-3\right)}}=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
23 tháng 12 2015
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)-\left(3x-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}+x-3\sqrt{x}-3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{-x+9}{x-9}=\frac{-\left(x-9\right)}{x-9}=-1\)
15 tháng 2 2020
Em mới học lớp 8 nhưng làm thử sai thì thôi nhé !!!
\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}\)\(-\frac{3x+3}{x-9}\)
\(P=\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\sqrt{x}^2-3^2}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\sqrt{x}^2-3^2}\)
\(p=\frac{-3\sqrt{x}-3}{\sqrt{x}^2-3^2}=\frac{-3.\left(\sqrt{x}+1\right)}{x-9}\)
9 tháng 7 2019
a) \(4x-\sqrt{x^2-4x+4}=4x-\sqrt{\left(x-2\right)^2}=4x-\left(x-2\right)=3x+2\)
b) \(3x+\sqrt{9+6x+x^2}=3x+\sqrt{\left(x+3\right)^2}=3x-\left(x+3\right)=2x-3\)
c) \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
d) \(\frac{\sqrt{x^2+4x+4}}{x+2}=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{\left|x+2\right|}{x+2}\)( 1 )
với x < -2 thì : \(\left(1\right)\Leftrightarrow\frac{-\left(x+2\right)}{x+2}=-1\)
với x > -2 thì : \(\left(1\right)\Leftrightarrow\frac{\left(x+2\right)}{x+2}=1\)
TH
13 tháng 10 2015
a/
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
b/ Vậy để P>1 khi BT trên>1
Ta có phương trình tương đương
\(x-3\sqrt{x}+3-x\sqrt{x}+6\text{x}-9>0\)
\(-x\sqrt{x}+7\text{x}-3\sqrt{x}-6>0\)
Giải pt rồi suy ra
tick cho mình nha
\(\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)