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a, \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-4^2\)
b, \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3+2z\right)\left(y-3-2z\right)=\left(y-3\right)^2-\left(2z\right)^2\)
c, \(\left(x-y-6\right)\left(x+y-6\right)=\left(x-6-y\right)\left(x-6+y\right)=\left(x-6\right)^2-y^2\)
d, \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z+x\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
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1.a) xy + 2y - x2 + 4
= y ( x + 2 ) - ( x2 - 4 ) = y ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 )( y - x + 2 )
b) 2x2 + y2 + 3xy
= ( 2x2 + 2xy ) + ( y2 + xy )
= 2x ( x + y ) + y ( x + y )
= ( x + y ) ( 2x + y )
2.
x - y = 5 \(\Rightarrow\)( x - y )2 = 25 \(\Rightarrow\)x2 + y2 = 25 + 2xy = 25 + 2.3 = 31
A = ( x + y )2 = x2 + y2 + 2xy = 31 + 6 = 37
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a) 6xy^3+x^2y^6+9
= (xy^3 + 3)^2
b) x^4-2x^2y+y^2
= (x^2 - y)^2
c) x^6+25-10x^3
= (x^3 - 5)^2
a/ 6xy3+x2y6+9
= (xy3+3)2 bình phương của 1 tổng;cttq: (A+B)2
b/ x4-2x2y+y2
= (x2-y)2 bình phương của 1 hiệu; cttq (A-B)2
c/ x6+25-10x3
=(x3-5)2
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a . \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-4^2\)
b . \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)
c . \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3\right)^2-\left(2z\right)^2\)
d . \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
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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
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`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`
`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`
`=-(x/2 - 1)^3`
`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`
`=(x^3 - 1/(2y))^{3}`
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\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
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1)a)x2+10x+26+y2+2y
=(x2+10x+25)+(y2+2y+1)
=(x+5)2+(y+1)2
b)x2-2xy+2y2+2y+1
=(x2-2xy+y2)+(y2+2y+1)
=(x-y)2+(y+1)2
c)z2-6z+13+t2+4t
=(z2-6z+9)+(t2+4t+4)
=(z-3)2+(t+2)2
d)4x2+2z2-4xz-2z+1
=(4x2-4xz+z2)+(z2-2z+1)
=(2x-z)2+(z-1)2
2)a)(x-3)2-4=0
<=>(x-3-2)(x-3+2)=0
<=>(x-5)(x-1)=0
<=>x-5=0 hoặc x-1=0
<=>x=5 hoặc x=1
b)x2-2x=24
<=>x2-2x-24=0
<=>(x2-6x)+(4x-24)=0
<=>x(x-6)+4(x-6)=0
<=>(x-6)(x+4)=0
<=>x-6=0 hoặc x+4=0
<=>x=6 hoặc x=-4
a) x^2 + 10x + 26 + y^2 + 2y
=x2+10x+25+y2+2y+1
=x2+2.x.5+52+y2+2.y.1+12
=(x+5)2+(y+1)2
b)x^2 - 2xy + 2y^2 + 2y +1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
c)z^2 - 6z + 13 + t^2 + 4t
=z2-6z+9+t2+4z+4
=z2-2.z.3+32+t2+2.t.2+22
=(z-3)2+(t+2)2
d)4x^2 + 2z^2 - 4xz - 2z + 1
=4x2-4xz+z2+z2-2z+1
=(2x)2-2.2x.z+z2+z2-2z.1+12
=(2x-z)2+(z-1)2
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`B=(x/2+y)^3-6(x/2+y)^2z + 6(x+2y)z^2-8z^3`
`=(x/2+y)^3 - 3. (x/2+y)^2 . 2z + 3. (x/2+y) . (2z)^2 - (2z)^3`
`=(x/2+y-2z)^3`
Sửa đề: Δ\(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)
Ta có: \(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)
\(=\left(\dfrac{1}{2}x+y\right)^2-3\cdot\left(\dfrac{1}{2}x+y\right)^2\cdot2z+3\cdot\left(\dfrac{1}{2}x+y\right)\cdot\left(2z\right)^2-\left(2z\right)^3\)
\(=\left(\dfrac{1}{2}x+y-2z\right)^3\)
a: \(x^3+\left(2y\right)^3=\left(x+2y\right)\left[x^2-x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b: \(\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
C.on b nha