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a)\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)
Vậy x=-1 y=1
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)
b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\) \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\)
\(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)
b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)
c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)
d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)
e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)
k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)
m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
e: \(49-x^2-2xy-y^2\)
\(=49-\left(x+y\right)^2\)
\(=\left(7-x-y\right)\left(7+x+y\right)\)
d: \(16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=0\\y=1\end{cases}\Rightarrow}x=-1}\)
Vậy x=-1 ; y=1
a: \(=\left(x+2-y\right)\left(x+2+y\right)\)
c: \(=\left(x-y\right)^2\)
\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)
Bài 1:
\(a,=\left(156-56\right)^2=100^2=10000\\ b,=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)\)
Bài 2:
\(a,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ b,=x^2+2x-6x-12=\left(x+2\right)\left(x-6\right)\\ c,=3\left(x^2-2xy-16+y^2\right)=3\left[\left(x-y\right)^2-16\right]\\ =3\left(x-y-4\right)\left(x-y+4\right)\)