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HH
11 tháng 6 2018
Giải:
a) \(\left(2x+y+3\right)^2\)
\(=\left(2x+y\right)^2+2.3\left(2x+y\right)+3^2\)
\(=\left(2x\right)^2+2.2x.y+y^2+2.3\left(2x+y\right)+3^2\)
\(=4x^2+4xy+y^2+12x+6y+9\)
Vậy ...
b) \(\left(x-2y+1\right)^2\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1^2\)
\(=x^2-2.x.2y+\left(2y\right)^2+2x-4y+1^2\)
\(=x^2-4xy+4y^2+2x-4y+1\)
Vậy ...
c) \(\left(x^2-2xy^2-3\right)^2\)
\(=\left(x^2-2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=\left(x^2\right)^2-2.x^2.2xy^2+\left(2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=x^4-4x^3y^2+4x^2y^4+6x^2-12xy^2-9\)
Vậy ...
YT
11 tháng 6 2018
\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)
19 tháng 6 2018
a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36y^2x^2-8y^3\)
c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)
HN
20 tháng 6 2018
a) \(\left(x^2+2xy\right)^3\)
\(=\left(x^2\right)^3+3\left(x^2\right)^22xy+3x^2\left(2xy\right)^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b) \(\left(3x^2-2y\right)^3\)
\(=\left(3x^2\right)^3-3\left(3x^2\right)^22y+3.3x^2\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
c) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\left(2x^3\right)^2y^2+3.2x^3\left(y^2\right)^2-\left(y^2\right)^3\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6.\)
5 tháng 11 2017
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
22 tháng 5 2022
a: \(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)
b: \(Q=x^2+4xy+4y^2-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
=25
22 tháng 5 2022
a: Sửa đề: y=2/3
\(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)
b: \(Q=x^2+4xy+4y^2-2x-4y+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
YP
2 tháng 10 2018
\(a,x^2+4y^2-4xy\)
\(\Rightarrow\)\(x^2-4xy+\left(2y\right)^2\)
\(\Rightarrow\)\(\left(x-2y\right)^2\)
BT
2 tháng 10 2018
a, x2 +4y2 -4xy = x2 - 4xy +4y2 = (x - 2y)2
b, x2 y4 +1 - 2xy2 - 9 = x2y4 - 2xy2 +1 -9 =( x2y4 -2xy2 +1)
= (xy2 -1 )2 - 9 =(xy2 -1+3)(xy2 - 1-3)
c, x2- 4x -3 = x2 - 4x +4 - 7
= ( x - 2)2 -7
d, C1 : x2 -8x + 7 = x2 -x -7x +7
= (x2 - x) - (7x +7)
= x(x-1) - 7(x-1)
= ( x - 1)(x - 7)
C2 : x2 - 8x + 7
= x2 - 8x + 16 - 9
= (x2 - 8x +16) -9
= (x - 4 )2 -9
= ( x - 4 +3 )(x - 4 -3 )
=( x - 1 ) (x - 7 )
Good luck !
Bn ko hiểu j cứ hỏi mik nhé ! 
\(x^2y^4+2xy^2+1=\left(xy^2\right)^2+2.xy^2.1+1^2=\left(xy^2+1\right)^2\)
Áp dụng hằng đẳng thức thứ nhất: \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(2xy^2+x^2y^4+1\)
\(=\left(xy^2\right)^2+2xy^2+1\)
\(=\left(xy^2+1\right)^2\)