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A=\(\frac{X^2-9-\left(4x-2\right).\left(x-3\right)}{x^2-6x+9}\)
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8 tháng 3 2020
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)
\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)
8 tháng 3 2020
\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)
\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)
\(\Leftrightarrow M=\frac{x-9}{2x}\)
Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)
HT
1
26 tháng 8 2017
\(\frac{x^2-9-\left(4x-2\right)\left(x-3\right)}{x^2-6x+9}\)
\(=\frac{\left(x-3\right)\left(x+3\right)-\left(4x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\frac{\left(x-3\right)\left[\left(x+3\right)-\left(4x-2\right)\right]}{\left(x-3\right)^2}\)
\(=\frac{x+3-4x+2}{x-3}\)
\(=\frac{-3x+5}{x-3}\)
26 tháng 8 2020
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
MN
5 tháng 3 2020
\(ĐKXĐ:x\ne\pm3\)
\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)
19 tháng 2 2019
a) \(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)(ĐKXĐ: \(\hept{\begin{cases}x\ne\pm2\\x\ne3\end{cases}}\))\(=\left[\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}\right]:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)\(=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{2-x}{x-3}=\frac{4x}{x-3}\)
b) l\(x-5\)l\(=2\Leftrightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\left(n\right)\\x=3\left(l\right)\end{cases}\Rightarrow A=\frac{4.7}{7-3}=\frac{28}{4}=7}\)
c)
* Để A có giá trị là một số nguyên thì \(A=\frac{4x}{x-3}=\frac{4x-12+12}{x-3}=4+\frac{12}{x-3}\)là một số nguyên hay \(\frac{12}{x-3}\)là một số nguyên \(\Rightarrow x-3\inƯ\left(12\right)\Rightarrow S=\left(-9;-3;-1;0;1;4;5;6;7;9;15\right)\)(1)
* Để \(A=4+\frac{12}{x-3}< 4\Leftrightarrow\frac{12}{x-3}< 0\) thì \(x-3< 0\Leftrightarrow x< 3\)(2)
(1)(2) \(\Rightarrow S=\left(-9;-3;-1;0;1\right)\)
20 tháng 8 2023
a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)
\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)
b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)
=>\(\frac{x^2-3^2-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x+3\right)\cdot\left(x-3\right)-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x-3\right)\cdot\left(x+3-4x+2\right)}{\left(x-3\right)^2}\)
=>\(\frac{-3x+5}{x-3}\)
cho minh nhe!