giải giúp tui bài này nha
tìm gtln: A=-2x2-y2+2xy+4x-40
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a: 5x-20xy
\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)
b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)
c: \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
=(x-y-z)(x-y+z)
d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)
e; \(x^2+4x+3=x^2+x+3x+3\)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f: \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left\lbrack\left(x+y\right)^2-1\right\rbrack\)
=(x+y)(x+y-1)(x+y+1)
g: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
=(x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
h: \(16x-5x^2-3\)
\(=-5x^2+15x+x-3\)
=-5x(x-3)+(x-3)
=(x-3)(-5x+1)
i: \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)
k: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\cdot\left(x-2\right)\left(x+2\right)\)
l: \(x^2-y^2-5x+5y\)
=(x-y)(x+y)-5(x-y)
=(x-y)(x+y-5)
\(-2x^2-2xy-y^2+2x-2y-2=-\left[y^2+2y\left(x+1\right)+\left(x+1\right)^2\right]-\left(x^2-4x+4\right)+3=-\left(y+x+1\right)^2-\left(x-2\right)^2+3\le3\)
\(max=3\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
Học tốt!
`2(x^2+y^2)+z^2=-2xy+2yz-4x-4`
`<=>2x^2+2y^2+z^2+2xy-2yz+4x+4=0`
`<=>(x^2+2xy+y^2)+(y^2-2yz+z^2)+(x^2+4x+4)=0`
`<=>(x+y)^2+(y-z)^2+(x+2)^2=0`
Vì `VT>=0`
Nên dấu "=" xảy ra khi `x+y=0,y-z=0,x+2=0`
`<=>x=-y,y=z,x=-2`
`<=>x=-2,y=z=-x=2`
Vậy `(x,y,z)=(-2,2,2)`
Ta có: \(P=2x-2xy-2x^2-y^2\)
\(P=-x^2-2xy-y^2-x^2+2x\)
\(P=-\left(x^2+2xy+y^2\right)-\left(x^2-2x+1\right)+1\)
\(P=-\left(x+y\right)^2-\left(x-1\right)^2+1\)
\(P=-\left[\left(x+y\right)^2+\left(x-1\right)^2\right]+1\le1\forall x;y\)
Vậy GTLN của P là 1 khi x=-1; y=1.
\(=2x^2\left(x-1\right)-4x\left(x-1\right)=\left(x-1\right)\left(2x^2-4x\right)=2x\left(x-2\right)\left(x-1\right)\)
\(=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right).\left(x-y+3\right)\)
A=\(-\left(2x^2+y^2-2xy-4x+40\right)=\)\(-\left[\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+36\right]\)=\(-\left[\left(x-y\right)^2+\left(x-2\right)^2+36\right]\)=\(-\left(x-y\right)^2-\left(x-2\right)^2-36\le-36\)
dấu "='' xảy ra \(\Leftrightarrow x=y=2\)