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Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
a, x\(^3\)+9x\(^2\)-4x-36
=x(x\(^2\)-4)+9(x\(^2\)-4)
=(x\(^2\)-4)(x+9)
=(x-2)(x+2)(x+9)
b,x\(^2\)-7x-10
=x\(^2\)-5x-2x-10
=(x-5)(x-2)
{câu này hình như sai đê bài hay sao ý}
a/x3+9x2-4x-36
=(x3-4x)+(9x2-36)
=x(x2-4)+9(x2-4)
= (x2-4)(x+9)
=(x-2)(x+2)(x+9)
b/3x2-7x-10
=3x2-10x+3x-10
=x(3x-10)+1(3x-10)
= (x+1)(3x-10)
2) a)\(p=\dfrac{x^3+2x^2+x}{x^3-x}=\dfrac{x\left(x^2+2x+1\right)}{x\left(x^2-1\right)}=\dfrac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
DKXD : x(x-1)(x+1)≠0
=> \(\left[{}\begin{matrix}x\ne0\\x-1\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne0\\x\ne1\\x\ne-1\end{matrix}\right.\)
Vay.......
b) \(p=\dfrac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
d, 3x\(^2\)-3y\(^2\)-12x+12y
=3(x\(^2\)-y\(^2\)-6x+6y)
=3[(x-y)(x+y)-6(x-y)]
=3(x-y)(x+y-6)
e,x\(^2\)-3x-2
=x\(^2\)-2x-x-2
=(x-2)(x-1)
f,x\(^3\)-4x\(^2\)+4x
=x(x\(^2\)-4x+4)
=x(x-2)\(^2\)
g,x\(^2\)-xy+7x-7y
=x(x-y)+7(x-y)
=(x-y)(x+7)
h,x\(^2\)-y\(^2\)+2x-2y
=(x-y)(x+y)+2(x-y)
(x-y)(x+y+2)
bài 2
a,để phân thức được xác định thì x\(^3\)-x\(\ne\)0
=>x(x-1)(x+1)\(\ne\)0
=>x\(\ne\)0,x\(\ne\pm\)1
b,với x\(\ne\)0;x \(\ne\pm\)1 thì
P=\(\dfrac{x^3+2x^2+x}{x^3-x}\)
P=\(\dfrac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
P=\(\dfrac{x+1}{x-1}\)
c,5x\(^3\)-x\(^2\)y-10x\(^2\)+10xy
=x(5x\(^2\)-xy-10x+10y)
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