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A=2(x+y)+3xy(x+y)+5x2y2(x+y)+2
A=2.0+3xy.0+5x2y2.0+2
A=2
B=xy(x+y)+2x2y (x+y)+5
B=xy.0+2x2y.0+5=5
a,Ta có 2(x+y)+3xy(x+y)+5x2y2(x+y)+4
Xg thay x+y=0 vào là dc bn nhó
Chúc bn hok tốt
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
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\(^2\)