tìm max 1)m=11-`10x-x2+2x+10y-8
2)n=2xy-x2-4y2
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a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)
\(=\left(x-y\right)^2-10\left(x-y\right)\)
\(=9^2-10\cdot9=-9\)
a.
$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.
$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.
$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$
d.
$x^3-3x^2+3x-1=(x-1)^3$
e.
$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$
$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$
f.
$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$
h: \(=\left(x+3\right)\cdot\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) Ta có: \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
Vậy: \(M=x^2+11xy-y^2\)
b) Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)
Vậy: \(N=-x^2+10xy-12y^2\)
a, (6x2+9xy-y2) - ( 5x2-2xy)=M
=> M= (6x2+9xy-y2) - ( 5x2-2xy)
=> M= 6x2+9xy-y2 - 5x2+2xy
=> M=(6x2- 5x2)+(9xy+2xy)-y2
=>M= 1x2 + 11xy - y2
Vậy M= 1x2 + 11xy - y2
b, N= (3xy-4y2) - (x2-7xy+8y2)
=> N= 3xy-4y2 - x2+7xy-8y2
=> N= (3xy+7xy)-(4y2+8y2)-x2
=> N= 10xy - 12y2 -x2
Vậy N= 10xy - 12y2 -x2
a: Ta có: \(M+5x^2-2xy=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
b: Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)