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\(=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\)
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
Ta có: \(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(x+3\right)\)
\(=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)
x2 + 4z2 - 4t2 - 4xt
= x2 - 4xt - 4t2 + 4z2
= 4t2 - 4xt + x2 + 4z2
= (2t - x)2 + 4z2
= \(-\left[\left(2t-x\right)^2-4z^2\right]\)
= \(-\left(2t-x-4z\right)\left(2t-x+4z\right)\)
Lm sao bn ra \(\left(2t-x\right)^2+4z^2=-\left[\left(2t-x\right)^2-4z^2\right]\) hay z?
\(x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+4=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)(vô lí)
Vậy pt vô nghiệm
= ( x2 - 4y2 ) - ( 2x + 4y )
= ( x - 2y ) ( x + 2y ) - 2 ( x - 2y )
= ( x - 2y ) ( x + 2y - 2 )
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
`x^2-2x-4y^2+4y`
`=(x^2-4y^2)-2x+4y`
`=(x-2y)(x+2y)-2(x-2y)`
`=(x-2y)(x+2y-2)`