9x:3x=27
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DT
26 tháng 6 2016
\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)
=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)
=\(\frac{x}{x-3}\)
DV
10 tháng 12 2016
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
MN
5 tháng 3 2020
\(ĐKXĐ:x\ne\pm3\)
\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)
PN
14 tháng 9 2015
(9x-63).(27-3x)=0(x> hoặc =7)
=> 9x-63=0 hoặc 27-3x=0
tính lần lượt ra loại 2-x vì x> hoặc =7 =>x=7
8 tháng 2 2022
ĐKXĐ: \(x\ne\pm3\)
\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
Ý 2 mình k hiểu ý bạn lắm
\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)
\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)
KK
28 tháng 10 2017
a) 9x (3x-y) +3y (y-3x)
= 9x(3x-y)-3y(3x-y)
=(3x-y)(9x-3y)
= (3x-y)3(3x-y)
= 3(3x-y)2
b)x3 - 3x2 - 9x + 27
=x3+3x2-6x2-18x+9x+27
= (x3+3x2)-(6x2+18x)+(9x+27)
= x2(x+3)-6x(x+3)+9(x+3)
= (x+3)(x2-6x+9)
=(x+3)(x-3)2
PT
28 tháng 10 2017
a) \(9x\left(3x-y\right)+3y\left(y-3x\right)=9x\left(3x-y\right)-3y\left(3x-y\right)\)
=\(\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(3x-y\right)\)
\(=3\left(3x-y\right)^2\)
b) \(x^3-3x^2-9x+27=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-9\right)\)
\(=\left(x-3\right)\left(x-3\right)\left(x+3\right)\)
=\(\left(x-3\right)^2\left(x+3\right)\)
21 tháng 10 2016
\(x^3-3x^2-9x+27\)
\(x^3+27-3x^2-9x\)
\(\left(x+3\right)\left(x^2-6x+9\right)-3x\left(x+3\right)\)
\(\left(x+3\right)\left(x^2-6x+9-3x\right)\)
\(\left(x+3\right)\left(x^2-9x+9\right)\)
\(9^3:3^3=27\)
Click cho mik nha !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(9^x:3^x=27\)
\(\Rightarrow\left(9:3\right)^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)