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b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
\(a.=\left(2x\right)^2-2.2x.2y+\left(2y\right)^2=\left(2x-2y\right)^2\)
\(b.=\left(3x\right)^2-2.3x.2+2^2=\left(3x-2\right)^2\)
a. 4x2+4y2-8xy=(2x)2+(2y)2-8xy
=(2x-2y)2
b.9x2-12x+4=(3x)2-12x+22
=(3x-2)2
c.xy2+1/4x2y4+1=xy2+(1/2xy2)2+1
=(1/2xy2+2)2
Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
Bài 1:
a) \(x^2\left(5x^3-x-6\right)\)
\(=x^2.5x^3-x^2.x-x^2.6\)
\(=5x^5-x^3-6x^2\)
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2xy\left(x-y\right)+y^2\left(x-y\right)\)
\(=x^3-x^2y-2x^2y+2xy^2+y^2x-y^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
Bài 2:
a) \(y^2+2y+1\)
\(=\left(y+1\right)^2\)
b) \(9x^2+y^2-6xy\)
\(=\left(3x\right)^2-2.3x.y+y^2\)
\(=\left(3x-y\right)^2\)
c) \(25a^2+4b^2+20ab\)
\(=\left(5a\right)^2+2.5a.2b+\left(2b\right)^2\)
\(=\left(5a+2b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) x^2 - x + 1/4
= x^2 - 2.x + 1/4 + (1/2)^2
= ( x - 1/2)^2
a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
a)\(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
b)\(9x^2+42x+49=\left(3x\right)^2+2\cdot3x\cdot7+7^2=\left(3x+7\right)^2\)
c)\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{1}{3}\cdot y^4+\left(y^4\right)^2=\left(y^4-\dfrac{1}{3}\right)^2\)
a) \(x^2+2.2x+2^2\)
\(=\left(x+2\right)^2\)
b)\(\left(3x\right)^2+2.3.7x+7^2\)
\(=\left(3x+7\right)^2\)
c) \(\left(\dfrac{1}{3}\right)^2-2.\dfrac{1}{3}.y^4+\left(y^4\right)^2\)
\(=\left(\dfrac{1}{3}-y^4\right)^2\)