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LL
1
7 tháng 7 2017
a. \(\left(2x^2+y\right)\left(2x^2-y\right)-4x^2+y^2=\left(2x^2\right)^2-y^2-4x^2+y^2\)
\(=4x^4-4x^2\)
b. \(\left(2x^2+y\right)^2-\left(2x^2-y^2\right)=4x^4+4x^2y+y^2-2x^2+y^2\)
\(=4x^4+4x^2y-2x^2+2y^2\)
c. \(\left(2x+1\right)\left(2x-1\right)-4x^2=\left(2x\right)^2-1^2-4x^2=4x^2-1-4x^2=-1\)
d. \(\left(2x^{3y}+y\right)^2-\left(y-2x^{3y}\right)^2\)
\(=\left(2x^{3y}+y+y-2x^{3y}\right)\left(2x^{3y}+y-y+2x^{3y}\right)\)
\(=2y.2.2x^{3y}=4y.2x^{3y}\)
26 tháng 6 2021
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
26 tháng 6 2021
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
AT
7 tháng 7 2017
a/ \(\left(2x+1\right)\left(2x-1\right)-4x^2=\left(2x\right)^2-1^2-4x^2\)
\(=4x^2-1-4x^2\)
b/ \(\left(2x^2+y\right)\left(2x^2-y\right)-4x^2+y^2\)
\(=\left(2x^2\right)^2-y^2-4x^2+y^2=4x^4-y^2-4x^2+y^2=4x^4-4x^2\)
c/ \(\left(2x^2+y\right)^2-\left(2x^2-y\right)^2\)
\(=\left(2x^2+y+2x^2-y\right)\left(2x^2+y-2x^2+y\right)\)
\(=4x^2\cdot2y=8x^2y\)
d/ \(\left(2x^3y+y\right)^2-\left(y-2x^3y\right)^2=\left(2x^3y+y\right)^2-\left(2x^3y-y\right)^2\)
\(=\left(2x^3y+y+2x^3y-y\right)\left(2x^3y+y-2x^3y+y\right)\)
\(=4x^3y\cdot2y=8x^3y^2\)
TH
1
BT
9 tháng 11 2016
(4x2+2xy+y2)(2x-y)-(2x-y)(4x2-2xy+y2)
=(2x3-y3)-(2x+y)(4x2-2xy+y2)
=(2x3-y3)-(2x3+y3)
=2x3-y3-2x3+y3
=0
LT
2
SM
23 tháng 9 2018
( 2x + y ) ( 4x2 - 2xy + y2 ) - ( 2x - y ) ( 4x2 + 2xy + y2 )
= 8x3 + y3 - ( 8x3 - y3 )
= 2y3
23 tháng 9 2018
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=\left(8x^3-8x^3\right)+\left(y^3+y^3\right)\)
\(=2y^3\)
HT
1
27 tháng 9 2017
a,=(x+2).(x^2-2x+2^2)-18-x^3
=x^3 + 2^3 - 18 -x^3=(x^3-x^3)+(8-18) = -10
b, =(2x-y).((2x)2+2xy +y2) - (2x +y).((2x)^2-2xy +y^2)
=(2x)3-y3- (2x)3-y3= -2 .y3
2 ý này áp dụng HĐT : x3+y3=(x+y).(x2-xy+y2)
x3-y3=(x-y).(x2+xy+y2)
31 tháng 1 2019
\(\left(\dfrac{2x}{2x+y}-\dfrac{4x^2}{4x^2+4xy+y^2}\right):\left(\dfrac{2x}{4x^2-y^2}+\dfrac{1}{y-2x}\right)\)
=\(\left[\dfrac{2x}{2x+y}-\dfrac{4x^2}{\left(2x+y\right)^2}\right]:\left[\dfrac{2x}{\left(2x-y\right)\left(2x+y\right)}-\dfrac{1}{2x-y}\right]\)
\(=\left[\dfrac{2x\left(2x+y\right)}{\left(2x+y\right)^2}-\dfrac{4x^2}{\left(2x+y\right)^2}\right]:\left[\dfrac{2x}{\left(2x-y\right)\left(2x+y\right)}-\dfrac{y+2x}{\left(2x-y\right)\left(y+2x\right)}\right]\)
\(=\left[\dfrac{4x^2+2xy-4x^2}{\left(2x+y\right)^2}\right]:[\dfrac{2x-y-2x}{\left(2x-y\right)\left(2x+y\right)}]\)
\(=\dfrac{2xy}{\left(2x+y\right)^2}:\dfrac{-y}{\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{2xy\left(2x-y\right)\left(2x+y\right)}{\left(2x+y\right)\left(2x+y\right)\left(-y\right)}\)
\(=\dfrac{2x\left(2x-y\right)}{-\left(2x+y\right)}\)
\(\dfrac{4x^2-2xy}{-2x-y}\)
\(\frac{4x^2-y^2}{2x^2+y}\)\(=\frac{\left(2x^2-y\right)\left(2x^2+y\right)}{2x^2+y}\)\(=2x^2-y\)