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=x2-2xy+1-4y2-4y-1
=(x-1)2-(4y2+4y+1)
=(x-1)2-(2y+1)2
=(x-1+2y+1)(x-1-2y-1)
=(x+2y)(x-2y-2)
\(x^4+3x^2y^2+4y^4\)
\(x^4+4y^4-2xy^3+2xy^3+2x^2y^2+2x^2y^2-x^2y^2\)
\(+x^3y-x^3y\)
\(=\left(4y^4-2xy^3+2x^2y^2\right)+\left(2xy^3-x^2y^2+x^3y\right)\)
\(+\left(2x^2y^2-x^3y+x^4\right)\)
\(=2y^2\left(2y^2-xy+x^2\right)+xy\left(2y^2-xy+x^2\right)\)
\(+x^2\left(2y^2-xy+x^2\right)\)
\(=\left(2y^2+xy+x^2\right)\left(2y^2-xy+x^2\right)\)
Answer:
\(25x^2-10x+4y-4y^2\)
\(=25x^2-10x+1-4x^2+4y-1\)
\(=\left(25x^2-10x+1\right)-\left(4y^2-2y+1\right)\)
\(=[\left(5x\right)^2-2.5x.1+1]-[\left(2y\right)^2-2.2y.1+1]\)
\(=\left(5x-1\right)^2-\left(2y-1\right)^2\)
\(=\left(5x-1-2y+1\right).\left(5x-1+2y-1\right)\)
\(=\left(5x-2y\right).\left(5x+2y-2\right)\)
a/ 16x2 - 4y2 = (4x + 2y)(4x - 2y)
b/ x4 + 1 = (x2 + √2 x + 1)(x2 - √2 x + 1)
a/ 16x2-4y2=(4x)2-(2y)2=(4x+2y)(4x-2y)
b/ x4+1=(x2)2+12=(x2+1)(x2-1)
k minh nha
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
a ) \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-2\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b ) \(x^4+2x^3-4x-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)
a,x2-2x-4y2-4y=(x2-4y2)-(2x+4y)
=(x-2y).(x+2y)-2(x+2y)
=(x+2y).(x-2y-2)
a) x2-2x-4y2-4y = (x2-4y2) -2(x+2y)= (x-2y)(x+2y) - 2(x+2y)= (x+2y)(x-2y-2)
b) x4+2x3-4x-4=(x2-2)(x2+2) +2x(x2-2)=(x2-2)(x2+2+2x)
NHớ chọn mik nha :)
9x2 - 9xy - 4y2
=( 9x2 - 4y2 ) - 9xy
= ( 3x - 2y ) ( 3x + 2y ) - 9xy
\(4y^4+1\\ =4y^4+4y^2+1-4y^2\\ =\left(4y^4+4y^2+1\right)-4y^2\\ =\left(2y^2+1\right)^2-\left(2y\right)^2\\ =\left(2y^2-2y+1\right)\left(2y^2+2y+1\right)\)