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Ta có: 2x^2+2y^2=x^2-2xy+y^2
<=>x^2+y^2=-2xy
<=> x^2+2xy+y^2=0
<=>(x+y)^2=0
=>x+y=0
=>x=-y
2(x2+y2)=(x-y)2 => 2x2+2y2=x2-2xy+y2
x2+y2=-2xy
x2+y2+2xy=0
=>(x+y)2=0 =>x+y=0 => x=-y
Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)
\(P-\dfrac{5}{2}=x+2y-\dfrac{x^2+y^2}{2}=-\dfrac{1}{2}\left(x-1\right)^2-\dfrac{1}{2}\left(y-2\right)^2+\dfrac{5}{2}\le\dfrac{5}{2}\)
\(\Rightarrow P-\dfrac{5}{2}\le\dfrac{5}{2}\Rightarrow P\le5\)
\(P_{max}=5\) khi \(\left(x;y\right)=\left(1;2\right)\)
`{((a-1)x+y=a),(x+(a-1)y=2):}`
`<=>{(ax-x+y=a),(x+ay-y=2):}`
`<=>{(a(x-1)=x-y<=>a=[x-y]/[x-1]),(x+[x-y]/[x-1]-y=2):}`
`<=>x(x-1)+x-y-y(x-1)=2(x-1)`
`<=>x^2-x+x-y-xy+y=2x-2`
`<=>x^2-xy-2x+2=0`
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`b)x^2-xy-2x+2=0`
`<=>xy=x^2-2x+2`
`<=>y=x-2+2/x`
Thay `y=x-2+2/x` vào `6x^2-17y=7` có:
`6x^2-17(x-2+2/x)=7`
`<=>6x^3-17x^2+34x-34-7x=0`
`<=>6x^3-12x^2-5x^2+10x+17x-34=0`
`<=>(x-2)(6x^2-5x+17)=0`
Mà `6x^2-5x+17 > 0`
`=>x-2=0<=>x=2`
`=>y=2-2+2/2=1`
Thay `x=2;y=1` vào `(a-1)x+y=a` có: `(a-1).2+1=a<=>a=1`
`(x-y+z)^2+(z-y)^2+2(x-y+z)(y-z)=(x-y+z+z-y)^2`
`=(x-2y+2z)^2`
`=x^2+4y^2+4z^2-4xy-8yz+4zx`
`=>` Hệ số của `x^2` là: `1`.