Thu gọn đa thức
D = \(-4\dfrac{1}{3}xy^3+6x^2y-3\dfrac{1}{2}x^3y-2xy^3+x^2y\)
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\(A=-2xy^2+xy^2+\dfrac{1}{3}x^3y-\dfrac{1}{3}x^3y-x+x-4x^2y=-xy^2-4x^2y\)
bậc là 3
Thu gọn đa thức:
\(C=-\dfrac{1}{2}x^2y-2xy+\dfrac{1}{2}x^2y-xy+xy-\dfrac{1}{3}x+\dfrac{1}{2}+x-0,25\)
\(=x^2y\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+xy\left(-2-1+1\right)+x\left(-\dfrac{1}{3}+1\right)+\dfrac{1}{2}-\dfrac{1}{4}\)
\(=-2xy+\dfrac{2}{3}x+\dfrac{1}{4}\)
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)
Bậc:3
Thay x=-1, y=1 vào B ta có:
\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
Vậy \(M=x^2+11xy-y^2\)
\(b.M+(3x^2y-2xy^3)=2x^2y-4xy^3\)
\(M=(2x^2y-4xy^3)-(3x^2-2xy^3)\)
\(M=
\) \(2x^2-4xy^3-3x^2+2xy^3\)
\(M=(2x^2-3x^2)+(-4xy^3+2xy^3)\)
\(M=-x^2-2xy^3\)
Vậy \(M=-x^2-2xy^3\)
a) M + (5x\(^2\) - 2xy) = 6x\(^2\) + 9xy - y\(^2\)
=> M = (6x\(^2\) + 9xy - y\(^2\)) - (5x\(^2\) - 2xy)
M = 6x\(^2\) + 9xy - y\(^2\) - 5x\(^2\) + 2xy
M = (6x\(^2\) - 5x\(^2\)) + (9xy + 2xy) - y\(^2\)
M = 1x\(^2\) + 11xy - y\(^2\)
\(A=x^3.\left(-\dfrac{5}{4}x^2y\right).\left(\dfrac{2}{5}x^3y^4\right).\\ A=-\dfrac{1}{2}x^8y^5.\)
- Bậc: 8.
- Hệ số: \(-\dfrac{1}{2}.\)
- Biến: \(x;y.\)
\(B=\left(-\dfrac{3}{4}x^5y^4\right).\left(xy^2\right).\left(-\dfrac{8}{9}x^2y^3\right).\\ B=\dfrac{2}{3}x^8y^9.\)
- Bậc: 9.
- Hệ số: \(\dfrac{2}{3}.\)
- Biến: \(x;y.\)
\(D=xy^3\left(-4-\dfrac{1}{3}-2\right)+x^2y\left(6+1\right)-\dfrac{7}{2}x^3y=-\dfrac{19}{3}xy^3+7x^2y-\dfrac{7}{2}x^3y\)