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Câu 1:\(x^2.y+x.y^2-x-y=x.\left(x.y-1\right)+y.\left(x.y-1\right)=\left(x+y\right).\left(x.y-1\right)\)
Câu 3:\(a.x^2+a.y-b.x^2-b.y=x^2.\left(a-b\right)+y.\left(a-b\right)=\left(x^2+y\right).\left(a-b\right)\)
`A=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2)`
`=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x`
`=(x^4-x^4)+(x^3-x^3)+(3x^2-x^2-2x^2)+(2x-2x)+2`
`=2`
a ) -36a2 + x2 + 4y2 - 4xy
= ( x2 - 4xy + 4y2 ) - (6a)2
= ( x -2y )2 - (6a)2
= ( x - 2y - 6a ).(x - 2y + 6a )
b ) 10ax - 5ay +2x - y
= ( 10ax - 5ay ) + ( 2x - y )
= 5a ( 2x - y ) + ( 2x - y )
= ( 2x - y ) . (5a + 1 )
c ) 2a2b(x + y) - 4ab2(-x - y )
= 2a2b( x+ y ) + 4ab2(x + y )
= 2ab(x + y ) ( a + 2b )
a, \(-36a^2+x^2+4y^2-4xy=\left(x+2y\right)^2-\left(6a\right)^2=\left(x+2y-6a\right)\left(x+2y+6a\right)\)
b, \(10ax-5ay+2x-y=5a\left(2x-y\right)+2x-y=\left(5a+1\right)\left(2x-y\right)\)
c, \(2a^2b\left(x+y\right)-4ab^2\left(-x-y\right)=2a^2b\left(x+y\right)+4ab^2\left(x+y\right)\)
\(=\left(2a^2b+4ab^2\right)\left(x+y\right)=2ab\left(a+2b\right)\left(x+y\right)\)
\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)
\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)
\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)
\(ĐK:x\ge2016;y\ge2017;z\ge2018\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)
1)M=3x(2x-5y)+(3x-y)(-2x)-1/2(2-26xy)
=-1
2)
a)7x(x-2)-5(x-1)=21x^2-14x^2+3
<=>7x2-19x+5=7x2+3
<=>-19x=-2
<=>x=\(\frac{2}{19}\)