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C= x2 y - \(\dfrac{1}{2}\)xy2 + \(\dfrac{1}{3}\)x2y +\(\dfrac{2}{3}\)xy2 + 1
C=(x2y + \(\dfrac{1}{3}\)x2y )+( - \(\dfrac{1}{2}\)xy2 +\(\dfrac{2}{3}\)xy2)+ 1
C=\(\dfrac{4}{3}\)x2y +\(\dfrac{1}{6}\)xy2+1
=>Bặc: 3
D= xy2z + 3xyz2 - \(\dfrac{1}{5}\)xy2z - \(\dfrac{1}{3}\)xyz2 - 2
D=(xy2z - \(\dfrac{1}{5}\)xy2z )+( 3xyz2 - \(\dfrac{1}{3}\)xyz2) - 2
D=\(\dfrac{4}{5}\)xy2z +\(\dfrac{8}{3}\)xyz2 - 2
=> Bậc :4
E = 3xy5 - x2y + 7xy - 3xy5 + 3x2y - \(\dfrac{1}{2}\)xy + 1
E=(3xy5- 3xy5) + (- x2y + 3x2y) + (7xy - \(\dfrac{1}{2}\)xy)+ 1
E= 2x2y + \(\dfrac{13}{2}\)xy + 1
=> Bậc: 3
K = 5x3 - 4x + 7x2 - 6x3 + 4x + 1
K= (5x3 - 6x3 ) + (- 4x + 4x) +1
K= -1x3 + 1
=>Bậc: 3
F = 12x3y2 - \(\dfrac{3}{7}\)x4y2 + 2xy3 - x3y2 + x4y2 - xy3 - 5
F=( 12x3y2 - x3y2) + (- \(\dfrac{3}{7}\)x4y2 + x4y2) + (2xy3 - xy3) -5
F=11x3y2 + \(\dfrac{4}{7}\)x4y2 + xy3 - 5
=> Bậc :6
CHÚC BN HỌC TỐT ^-^
1, \(3xy^2+Q=-7xy^2\)
\(\Rightarrow Q=-7xy^2-3xy^2\)
\(\Rightarrow Q=-10xy^2\)
2. \(P=3+5x^2-3xy+5y-5x^2-11+2xy+x^3\)
\(\Rightarrow P=\left(5x^2-5x^2\right)+\left(-3xy+2xy\right)+5y+x^3+\left(3-11\right)\)
\(\Rightarrow P=-xy+5y+x^3-8\)
Ta có: P(x)+Q(x)= x2y-2xy2+5xy+3+3xy2+5x2y-7xy+2=6x2y+xy2-2xy+5
a, \(M=x^2y+\frac{1}{3}xy^2+\frac{3}{5}xy^2-2xy+3x^2y-\frac{2}{3}\)
\(M=\left(x^2y+3x^2y\right)+\left(\frac{1}{3}xy^2+\frac{3}{5}xy^2\right)-2xy-\frac{2}{3}\)
\(M=4x^2y+\frac{8}{15}xy^2-2xy-\frac{2}{3}\)
b, Giá trị của biểu thức \(M=4x^2y+\frac{8}{15}xy^2-2xy-\frac{2}{3}\) tại \(x=-1\) và \(y=\frac{1}{2}\)
\(M=4.\left(-1\right)^2.\frac{1}{2}+\frac{8}{15}.\left(-1\right).\left(\frac{1}{2}\right)^2-2.\left(-1\right).\frac{1}{2}-\frac{2}{3}\)
\(M=4.1.\frac{1}{2}+\frac{8}{15}.\left(-1\right).\left(\frac{1}{4}\right)+1-\frac{2}{3}\)
\(M=2-\frac{2}{15}+1-\frac{2}{3}\)
\(M=\left(2+1\right)+\left(-\frac{2}{15}-\frac{2}{3}\right)\)
\(M=3+\left(\frac{-4}{5}\right)\)
\(M=\frac{11}{5}\)
Vậy giá trị của biểu thức \(M=4x^2y+\frac{8}{15}xy^2-2xy-\frac{2}{3}\) tại \(x=-1\) và \(y=\frac{1}{2}\) bằng \(\frac{11}{5}\)
\(=6x^3y^3-x^4y-\dfrac{7}{2}xy^2-5x^4y-6x^3y^3\)
\(=-\dfrac{7}{2}xy^2-6x^4y\)
a, \(M-\left(3xy-4y^2-2xy\right)=\left(x^2-7xy+8y^2\right)\)
\(\Rightarrow M=\left(x^2-7xy+8y^2\right)+\left(3xy-4y^2-2xy\right)\)
\(\Rightarrow M=x^2-7xy+8y^2+3xy-4y^2-2xy\)
\(\Rightarrow M=x^2+\left[3xy-7xy-2xy\right]+\left[8y^2-4y^2\right]\)
\(\Rightarrow M=x^2-6xy+4y^2\)
b, \(N+\left(x^3-xyz+3x^2y\right)=2x^3+3xy-xy^2\)
\(\Rightarrow N=\left(2x^3+3xy-xy^2\right)-\left(x^3-xyz+3x^2y\right)\)
\(\Rightarrow N=2x^3+3xy-xy^2-x^3+xyz-3x^2y\)
\(\Rightarrow N=\left[2x^3-x^3\right]+3xy-xy^2+xyz-3x^2y\)
\(\Rightarrow N=x^3+3xy-xy^2+xyz-3x^2y\)
Tích mình nha!!!
\(P=7xy^3+2xy^3-xy^3=\left(7+2-1\right)xy^3=8xy^3\\ Q=3xy-x^2+5xy^3-15xy-y^3=\left(3xy-15xy\right)-x^2+5xy^3-y^3=-12xy-x^2+5xy^3-y^3\)