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12 tháng 10

`a, 3x(1 + 4y) - 5y(1 - 3x)`
`= 3x + 12xy - 5y + 15xy`
`= 3x + (12xy + 15xy) - 5y`
`= 3x + 27xy - 5y`
`b, -3xy(4  -2xy+4x)`
`= -3xy . 4 + 3xy . 2xy - 3xy . 4x`
`= -12xy + 6x^2y^2 - 12x^2y`
`c, (4-2x)(12-4x-6y^2)`
`= 48-16x - 24y^2 - 24x + 8x^2 + 12xy^2`
`= 48+ (-16x - 24x) - 24y^2 + 8x^2 + 12xy^2`
`= 48 - 40x + 24y^2 + 8x^2 + 12xy^2`

 

12 tháng 10

`d, (5 + 2y)(-4x - 6xy - 7x)`
`= 5 . (-4x) - 5 . 6xy - 5 . 7x - 2y . 4x - 2y . 6xy - 2y . 7x`
`= -20x - 30xy - 35x - 8xy - 12xy^2 -14x`
`= (-20x - 35x - 14x) + (-30xy - 8xy) - 12xy^2`
`= -69x - 38xy - 12xy^2`
`e, (4xy^3 - 6xy^2 - 12xy^4)/(-2xy)`
`= 4xy^3 : (-2xy) + 6xy^2 : 2xy + 12xy^4 : 2xy`
`= -2y^2 + 3y + 6y^3`

 

24 tháng 8 2023

Để tính các biểu thức trên, ta sẽ áp dụng quy tắc nhân đa thức.

a) 2xy(3x+1) = 6x^2y + 2xy

b) -6x^2y(4x-5) = -24x^3y + 30x^2y

c) -3x^2(4x^2y-6xy) = -12x^4y + 18x^3y

d) 1/2xy^2(2x+3) = xy^2 + 3/2xy^2

e) 8x^2y^2(1/4xy-1/2x^2) = 2xy - 4x^2y^2

f) 5x(x^2+3x+1) = 5x^3 + 15x^2 + 5x

g) -1/2x^2y(2xy+6) = -x^3y - 3x^2y

5 tháng 10 2021

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

5 tháng 10 2021

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

 

1 tháng 9 2023

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

c) \(x^2+1+2x-y^2\)

\(=\left(x^2+2x+1\right)-y^2\)

\(=\left(x+1\right)^2-y^2\)

\(=\left(x+1+y\right)\left(x+1-y\right)\)

d) \(x^2+4x-2xy-4y+y^2\)

\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)

\(=\left(x-y\right)^2+4\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

\(=x\left(x^2-2x+1\right)\)

\(=x\left(x-1\right)^2\)

f) \(2x^2+4x+2-2y^2\)

\(=2\left(x^2+2x+1-y^2\right)\)

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

\(=2\left[\left(x+1\right)^2-y^2\right]\)

\(=2\left(x-y+1\right)\left(x+y+1\right)\)

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

17 tháng 8 2021

đề bài là rút gọn à

24 tháng 8 2023

a) (2xy+5)(4x^2+5): = 2xy * 4x^2 + 2xy * 5 + 5 * 4x^2 + 5 * 5 = 8x^3y + 10xy + 20x^2 + 25 b) (6xy+4)(2x^2+1): = 6xy * 2x^2 + 6xy * 1 + 4 * 2x^2 + 4 * 1 = 12x^3y + 6xy + 8x^2 + 4 c) (9x^2+4)(3x+5): = 9x^2 * 3x + 9x^2 * 5 + 4 * 3x + 4 * 5 = 27x^3 + 45x^2 + 12x + 20 d) (-2xy+6)(1/2xy+7): = -2xy * 1/2xy + (-2xy) * 7 + 6 * 1/2xy + 6 * 7 = -xy + (-14xy) + 3 + 42 = -15xy + 45 e) (4x+1)(2x^2+5x+2): = 4x * 2x^2 + 4x * 5x + 4x * 2 + 1 * 2x^2 + 1 * 5x + 1 * 2 = 8x^3 + 20x^2 + 8x + 2x^2 + 5x + 2 = 8x^3 + 22x^2 + 13x + 2 f) (2x^2y+3x)(2x+1): = 2x^2y * 2x + 2x^2y * 1 + 3x * 2x + 3x * 1 = 4x^3y + 2x^2y + 6x^2 + 3x g) (4xy+5x^2y)(2xy+6): = 4xy * 2xy + 4xy * 6 + 5x^2y * 2xy + 5x^2y * 6 = 8x^2y^2 + 24xy + 10x^3y + 30x^2y = 8x^2y^2 + 30x^2y + 24xy h) (-1/2x^2+6)(4xy+5): = -1/2x^2 * 4xy + (-1/2x^2) * 5 + 6 * 4xy + 6 * 5 = -2xy + (-5/2x^2) + 24xy + 30 = 22xy + (-5/2x^2) + 30

13 tháng 9 2023

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

\(=\left(x+y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3\)

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^3\)

\(=x^3-3x^2y+3xy^2-y^3\)

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3-y^3\)

\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

29 tháng 8 2023

\(a,3x\left(3x+6\right)=9x^2+18x\)

\(b,-\dfrac{1}{2}xy\left(4x^2+6x\right)\)

\(=-2x^3y-3x^2y\)

\(c,-2x^2y^3\left(\dfrac{1}{2}xy+4y^2\right)\)

\(=-x^3y^4-8x^2y^5\)

\(d,-6x^2\left(\dfrac{1}{3}xy^2-\dfrac{1}{2}y\right)\)

\(=-2x^3y^2+3x^2y\)

#\(Urushi\)

29 tháng 8 2023

a) 3x(3x+6) = 9x^2 + 18x b) -1/2xy(4x^2+6x) = -2x^3y - 3xy c) -2x^2y^3(1/2xy+4y^2 ) = -x^2y^2 - 8x^2y^5 d) -6x^2(1/3xy^2-1/2y) = -2xy + 3x^2y

Vậy kết quả của các biểu thức là: a) 9x^2 + 18x b) -2x^3y - 3xy c) -x^2y^2 - 8x^2y^5 d) -2xy + 3x^2y