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\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
Ta có : x4 - y4
= (x2)2 - (y2)2
= (x2 - y2)(x2 + y2)
= (x - y)(x + y)(x2 + y2)
b) 9(x - y)2 - 4(x + y)2
= [3(x - y) - 4(x + y)][3(x - y) + 4(x + y)]
= [3x - 3y - 4x - 4y][3x - 3y + 4x + 4y]
= (-x - 7y)(x + y)
e.\(x^4+2x^2+1=\left(x^2+1\right)^2\)
c.\(x^2-9y^2=\left(x-3y\right)\left(x+3y\right)\)
f.\(-x^2-2xy-y^2+1=-\left[\left(x+y\right)^2-1\right]=-\left(x+y-1\right)\left(x+y+1\right)=\left(x-y+1\right)\left(x+y+1\right)\)
g.\(x^3-x^2-x+1==x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)=\left(x-1\right)^2\left(x+1\right)\)
h.\(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
i.\(\left(x+y\right)^3-x^3-y^3=\left(x+y\right)^3-\left(x^3+y^3\right)=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
tíck mình nha bn thanks !!!!!
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
$=4x^2-4x+1-2(4x^2-12x+9)+4$
$=4x^2-4x+1-8x^2+24x-18+4$
$=-4x^2+20x-13$
b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
c) $(x^2+2xy)^2+2(x^2+2xy)y^2+y^4$$=(x^2+2xy)^2+2(x^2+2xy)y^2+(y^2)^2$
$=(x^2+2xy+y^2)^2$
d) $(x-1)^3+3x(x-1)^2+3x^2(x-1)+x^3$
Đặt $a=x-1,\ b=x$:
$=a^3+3a^2b+3ab^2+b^3$
$=(a+b)^3$
$=[(x-1)+x]^3$
$=(2x-1)^3$
e) $(2x+3y)(4x^2-6xy+9y^2)$Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:
$=(2x)^3+(3y)^3$
$=8x^3+27y^3$
f) $(x-y)(x^2+xy+y^2)-(x+y)(x^2-xy+y^2)$$=x^3-y^3-(x^3+y^3)$
$=-2y^3$
g) $(x^2-2y)(x^4+2x^2y+4y^2)-x^3(x-y)(x^2+xy+y^2)+8y^3$Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:
$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$
$=x^6-8y^3-x^6+x^3y^3+8y^3$
$=x^3y^3$