A+(2X^2-Y^5)=5X^2-3X^2+2XY
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
a: =18x^3y^2-12x^3y^3+6x^2y^2
b: (-3x+2)(5x^2-1/3x+4)
=-12x^3+x^2-12x+10x^2-2/3x+8
=-12x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)
=12x+34-x^3-12x+x^2+12
=-x^3+x^2+46
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
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
a, \(2x-5xy+3x^2\)Bậc : 2
b, \(ax^3+2xy-5\)Bậc : 3
c, \(5x^3-4x+7x^2-8x^3+4x+1-5x^2=-3x^3+2x^2+1\)Bậc : 3
d, \(-3x^5-x^3y-xy^2+3x^5+2=-x^3y-xy^2+2\)Bậc : 4
\(a. 2x(3x^2-5x+3) = 6x^3-10x^2+6x \)
\(b. -2x(x^2+5x-3) = -2x^3-10x^2+6x\)
c. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)
=-x^5+2x^3-\dfrac{3}{2}x^2\)
\(d.\left(2x-1\right)\left(x^2+5-4\right)=\left(2x-1\right)\left(x^2+1\right)=2x^3+2x-x^2-1\)
e. \(-\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=-10x^2+7x-12\)
f.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=\left(2x-y\right)\left(2x-y\right)^2=\left(2x-y\right)^3\)
g.\(\left(3x-4\right)\left(x+4\right)+\left(5-x\right)\left(2x^2+3x-1\right)=3x^2+12x-4x-16+10x^2+15x-5-2x^3-3x^2+x=-2x^3+10x^2+24x-21\)
e. \(7x\left(x-4\right)-\left(7x+3\right)\left(2x^2-x+4\right)=7x^2-28x-14x^3+7x^2-28x-6x^2+3x+-12=-14x^3+8x^2-53x-12\)
e: \(49-x^2-2xy-y^2\)
\(=49-\left(x+y\right)^2\)
\(=\left(7-x-y\right)\left(7+x+y\right)\)
d: \(16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
A + ( 2x2 - y5) = 5x2 - 3x2 + 2xy
=> A + ( 2x2 - y5) = 2x2 +2xy
=> A = 2x2 +2xy - ( 2x2 - y5)
=> A = 2x2 +2xy - 2x2 + y5
=> A = 2xy + y5
A+(2X2-Y5)=5X2-3X2+2XY
A + (2X2-Y5)= 2x2 + 2xy
A = 2x2 + 2xy - (2x2-y5)
A = 2x2 + 2xy -2x2 + y5
A = 2xy + y5