chịu rùi

bài này khó quá nguyen truong giang

chúc bn học tốt 

nhae$

hihi

x2 + y2 - x2y2 + xy - x - y
=(x2-x2y2)+(y2-y)+(xy-x) 
=x2(1-y)(1+y)-y(1-y)-x(1-y)
=(1-y)(x2+x2y-x-y)
=(1-y)[(x2-y)+(x2-x)]
=(1-y)[y(x-1)(x+1)+x(x-1)]
=(1-y)(x-1)(xy+x+y)

x2 + y2 - x2y2 + xy - x - y = (x2-x) + (y2-y) + (-x2y2 + xy) = x(x+1) + y(y+1) + xy(xy+1) = ( x+ y+ xy)( x + 1 + y + 1 + xy + 1)

\(x^2+y^2-x^2y^2+xy-x-y\)

\(=\left(x^2-x\right)+\left(y^2-y\right)+ \left(-x^2y^2+xy\right)\)

\(=x\left(x+1\right)+y\left(y+1\right)+xy\left(xy+1\right)\)

\(=\left(x+y+xy\right)\left(x+1+y+1+xy+1\right)\)

a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)

\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)

a) Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)

\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)

\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

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:

\(x^3+x^2+y^3+xy\)

\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)

C1:x^2+4y^2+4xy-16

=[x^2+4xy+(2y)^2]-16

=(x+2y)^2-4^2

=(x+2y-4)(x+2y+4)

C2: x^3+x^2+y^3+xy

=(x^2+xy)+(x^3+y^3)

=x(x+y)+(x+y)(x^2-xy+y^2)

=(x+y)(x+x^2-xy+y^2)

bài này ra lâu r nhưng ngứa tay nên giải luôn=)))))

đíu bt làm nka

\(x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+4=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)(vô lí)

Vậy pt vô nghiệm 

\(x^2+x+4=\left(x^2+4x+4\right)-3x=\left(x+2\right)^2-3x=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)\)

Giải thích: Sử dụng hằng đẳng thức 1 và 3 

⚽

\(x^2+y^2-x^2y^2+xy-x-y\)

\(=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)

\(=x^2\left(y-1\right)\left(-1-y\right)+y\left(y-1\right)+x\left(y-1\right)\)

\(=\left(y-1\right)\left(-x^2-x^2y+y+x\right)\)

\(=\left(y-1\right)\left[-x\left(x-1\right)-y\left(x-1\right)\left(x+1\right)\right]\)

\(=\left(y-1\right)\left(x-1\right)\left(-x-xy-y\right)\)

x²-x-y²-y=(x²-y²)-(x+y)=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

x²-xy+x-y=x(x-y)+(x-y)=(x+1)(x-y)

\(=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

\(x^2+5x+4\)

\(=\left(x^2+x\right)+\left(4x+4\right)\)

\(=x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x+4\right)\)

Ta có: \(x^2-2x-15\)

\(=x^2-5x+3x-15\)

\(=x\left(x-5\right)+3\left(x-5\right)\)

\(=\left(x-5\right)\left(x+3\right)\)

x² + 4z² - 4t² - 4xt

= x² - 4xt - 4t² + 4z²

= 4t² - 4xt + x² + 4z²

= (2t - x)² + 4z²

= \(-\left[\left(2t-x\right)^2-4z^2\right]\)

= \(-\left(2t-x-4z\right)\left(2t-x+4z\right)\)