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\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)
\(=3xy\left(x^2-2x-y^2-2yz-z^2+1\right)\)
\(=3xy\left[\left(x-1\right)^2-\left(y+z\right)^2\right]\)
\(=3xy\left(x-1-y-z\right)\left(x-1+y+z\right)\)
M = 6x2+3xy- 2y2- 5 +3y2 - 2x2-3xy+5
= (6x2- 2x2) + ( 3xy -3xy) + ( - 2y2- 2y2)+ (- 5+5)
= 4x2+ y2
Mà 4x2 >0
y2> 0
Vậy....
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=x^2+11xy-y^2\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=-x^2+10xy-12y^2\)
a. \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
b. \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
Bài 1:
\(A+B=7x^2-3xy+2y^2\)
\(A-B=x^2-7xy+4y^2\)
Bài 2:
a) \(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=x^2+11xy-y^2\)
b) \(N=\left(3xy-4y^2\right)-\left(x^2-7xy+8y^2\right)\)
\(N=-x^2-12y^2+10xy\)
a) Có:
\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6x^2-5x^2\right)+\left(9xy+2xy\right)-y^2\)
\(\Rightarrow M=x^2+11xy-y^2\)
b) Có:
\(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-\left(x^2-7xy+8y^2\right)\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=\left(3xy+7xy\right)+\left(-4y^2-8y^2\right)-x^2\)
\(\Rightarrow N=10xy+\left(-12y^2\right)-x^2\)
Hay \(N=10xy-12y^2-x^2\)
Chúc bạn học tốt!
\(P=\left(6xy-3y\right)-\left(-3xy+6x-8y\right)=6xy-3y+3xy-6x+8y\)
\(P=\left(6xy+3xy\right)+\left(-3y+8y\right)-6x\)
\(P=9xy+5y-6x\)
\(P=6xy-3y+3xy-6x+8y\)
\(=9xy+5y-6x\)