x^2-9y^2+x+3y
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PU
0
MP
2
KN
8 tháng 8 2019
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)+\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(=x^3-27y^3+x^3+27y^3=2x^3=2.\left(-1\right)^3=-2\)
NK
1
KT
10 tháng 8 2018
đk: \(x\ne0\); \(x\ne\pm3y\)
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x+3y}{x\left(x-3y\right)}\)
CN
24 tháng 12 2017
\(\dfrac{x+9y}{x^2-9y^2}-\dfrac{3y}{x^2+3xy}\)
\(=\dfrac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\dfrac{3y}{x\left(x+3y\right)}\)
\(=\dfrac{x\left(x+9y\right)-3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x-3y}{x\left(x+3y\right)}\)
NL
1
27 tháng 11 2018
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
\(=\frac{x\left(x+9y\right)-3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x+3y}{x\left(x-3y\right)}\)
T
11 tháng 12 2023
\(A=(x+3y)(x^2-3xy+9y^2)+3y(x+3y)(x-3y)-x(3xy+x^2-5)-5x+1\\A=(x+3y)[x^2-x\cdot3y+(3y)^2]+3y[x^2-(3y)^2]-3x^2y-x^3+5x-5x+1\\A=x^3+(3y)^3+3y(x^2-9y^2)-3x^2y-x^3+1\\A=x^3+27y^3+3x^2y-27y^3-3x^2y-x^3+1\\A=1\)$\Rightarrow$ Giá trị của $A$ không phụ thuộc vào giá trị của biến.
17 tháng 8 2021
a: \(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)=\left(2x+3y\right)^2:\left(2x+3y\right)=2x+3y\)
d: \(\left(x^2+6xy+9y^2\right):\left(x+3y\right)=\left(x+3y\right)^2:\left(x+3y\right)=x+3y\)
e: \(\dfrac{64y^3-27}{4y-3}=\dfrac{\left(4y-3\right)\left(16y^2+12y+9\right)}{4y-3}=16y^2+12y+9\)
17 tháng 8 2021
a, \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)
\(=\left(2x+3y\right)^2:\left(2x+3y\right)\\ =2x+3y\)
b,\(x^2+6xy+9y^2=\left(x+3y\right)^2\)
\(\Rightarrow\left(x^2+6xy+9y^2\right):\left(x+3y\right)\\ =\left(x+3y\right)^2:\left(x+3y\right)\\ =x+3y\)
c, \(64y^3-27=\left(4y-3\right)\left(16y^2+12y+9\right)\)
\(\Rightarrow\left(64x^3-27\right):\left(4y-3\right)\\ =\left[\left(4y-3\right)\left(16x^2+12x+9\right)\right]:\left(4y-3\right)\\ =16x^2+12x+9\)
2 tháng 9 2021
\(B=\left(x+3y\right)\left(x-3y\right)-y\left(x+9y\right)\)
\(=x^2-9y^2-xy-9y^2\)
\(=x^2-xy\)
\(C=\left(3x-9\right)\left(x^2+3x+9\right)-3x\left(x^2-2\right)\)
\(=3x^3-81-3x^3+6x\)
=6x-81
1)2xy−2y²+3x−3y1)2xy−2y²+3x−3y
=2y(x−y)+3(x−y)=2y(x−y)+3(x−y)
=(2y+3)(x−y)=(2y+3)(x−y)
2)x²−9y²+x+3y2)x²−9y²+x+3y
=x²−(3y)²+(x+3y)=x²−(3y)²+(x+3y)
=(x−3y)(x+3y)+(x+3y)=(x−3y)(x+3y)+(x+3y)
=(x−3y+1)(x+3y)=(x−3y+1)(x+3y)
3)3x(2x−3)−6x+93)3x(2x−3)−6x+9
=3x(2x−3)−3(2x−3)=3x(2x−3)−3(2x−3)
=(3x−3)(2x−3)=(3x−3)(2x−3)
=3(x−1)(2x−3)=3(x−1)(2x−3)
4)y²−25−x²−10x4)y²−25−x²−10x
=−(x²+10x+25)+y²=−(x²+10x+25)+y²
=−(x+5)²+y²=−(x+5)²+y²
=y²−(x+5)²=y²−(x+5)²
=(y−x−5)(y+x+5)=(y−x−5)(y+x+5)
5)x²−10x+245)x²−10x+24
=x²−4x−6x+24=x²−4x−6x+24
=x(x−4)−6(x−4)=x(x−4)−6(x−4)
=(x−6)(x−4)