\(D=xy^3\left(-4-\dfrac{1}{3}-2\right)+x^2y\left(6+1\right)-\dfrac{7}{2}x^3y=-\dfrac{19}{3}xy^3+7x^2y-\dfrac{7}{2}x^3y\)

\(A=-2xy^2+xy^2+\dfrac{1}{3}x^3y-\dfrac{1}{3}x^3y-x+x-4x^2y=-xy^2-4x^2y\)

bậc là 3

\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)

Bậc:3

Thay x=-1, y=1 vào B ta có:

\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)

\(B=x^5y^2+\dfrac{1}{2}x^5y^2-6xy+1=\dfrac{3}{2}x^5y^2-6xy+1\)

\(B=-\dfrac{1}{7}x^2y+x^5y^2-xy+\dfrac{1}{2}x^5y^2-5xy+\dfrac{1}{7}x^2y+2021^0\\ =\left(-\dfrac{1}{7}x^2y+\dfrac{1}{7}x^2y\right)+\left(x^5y^2+\dfrac{1}{2}x^5y^2\right)-\left(xy+5xy\right)+1\\ =0+\dfrac{3}{2}x^5y^2-6xy+1\\ =\dfrac{3}{2}x^5y^2-6xy+1\)

Thu gọn rồi tính giá trị của đa thức P tại x = 0,5 và y = 1.

Ta có: P = x² y + xy² – xy + xy² – 5xy – x²y

P = x² y – x²y + xy² + xy² – xy – 5xy = xy² – 6xy

Thay x = 0,5 và y = 1 ta được

P = . 0,5 . 1² – 6. 0,5 . 1 = - 3 = .

Vậy P = tại x = 0,5 và y = 1.

\(N=\dfrac{2}{5}x^2y+xy^2-3xy+\dfrac{1}{3}xy^2-3xy-\dfrac{1}{2}x^2y\)

\(=\left(\dfrac{2}{5}x^2y-\dfrac{1}{2}x^2y\right)+\left(xy^2+\dfrac{1}{3}xy^2\right)+\left(-3xy-3xy\right)\)

\(=-\dfrac{1}{10}x^2y+\dfrac{4}{3}xy^2-6xy\)

\(=-\dfrac{1}{10}.\left(0,5\right)^2.\left(-1\right)+\dfrac{4}{3}.0,5.\left(-1\right)^2-6.0,5.\left(-1\right)\)

\(=\dfrac{1}{40}+\dfrac{2}{3}+3=\dfrac{443}{120}\)

trong mat phang oxy cho tam giac ABC có C 9-2;-5/3),cos BC=4/5,Mthuoc BC,ME vuong goc AB,MF vuong goc AC,I(7/3;1/3) la trung diem AM.tim toa do A biet ym<0

1.

a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(\dfrac{1}{3}x^6y^5z\)

Deg=12

Mấy câu kia tương tự nha cố gắng lên!

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$