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MN
5 tháng 3 2020
\(ĐKXĐ:x\ne\pm3\)
\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)
TN
8 tháng 6 2016
A = (3x + y)^2 - 3y . ( 2x - 1/3y )
=2y2+9x2
B = ( x - 2 )^2 + ( x + 2 )^2 - 2. ( x - 2 ) ( x + 2)
=24
C = ( x - y ) ( x^2 + xy + y^2 ) + 2y^3
=y3+x3
D = ( x -5 ) ( x+ 5 ) -(x - 8 ) (x + 4)
=4x+7
E = (3x + 1 )^2 - 2 . ( 9x^2 - 1 ) + ( 3x - 1 ) ^2
=4
F = ( x - 3 ) ( x + 3 ) - ( x - 3 )^2
=6x-18
HD
25 tháng 7 2021
a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`
`=x^3-3^3-(9x^3+27x^2-9x-27)`
`=x^3-3^3-9x^3-27x^2+9x+27`
`=-8x^3-27x^2+9x`
b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`
`=x^3-2^3-x(x^2-9)`
`=x^3-8-x^3+9x`
`=9x-8`
25 tháng 7 2021
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)
\(=x^3-27-\left(9x^3+27x^2-9x-27\right)\)
\(=x^3-27-9x^3-27x^2+9x+27\)
\(=-8x^3-27x^2+9x\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=x^3-8-x\left(x^2-9\right)\)
\(=x^3-8-x^3+9x\)
\(=9x-8\)
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 \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)
LT
0
\(A=\left(x-2\right)^3-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(A=x^3-6x^2+12x-8-x^2+x^2-3x+3x^2-9x+8\)
\(A=x^3-6x^2+12x-3x+3x^2-9x\)
\(A=\left(-6x^2+3x^2\right)+\left(12x-3x-9x\right)+x^3\)
\(A=-3x^2+x^3\)