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\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=t\), ta có:
\(A=t^2-14t+24\)
\(=t^2-2t-12t+24\)
\(=t\left(t-2\right)-12\left(t-2\right)\)
\(=\left(t-2\right)\left(t-12\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(B=\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=t\), ta có:
\(B=t^2+4t-12\)
\(=t^2+6t-2t-12\)
\(=t\left(t+6\right)-2\left(t+6\right)\)
\(=\left(t+6\right)\left(t-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+4=t\), ta có:
\(C=t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
\(D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta có:
\(D=t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\), ta có:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) x2 + 2x - 3 = x2 - x + 3x - 3 = x( x - 1 ) + 3( x - 1 ) = ( x - 1 )( x + 3 )
b) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
c) x2 - 2x - 48 = x2 + 6x - 8x - 48 = x( x + 6 ) - 8( x + 6 ) = ( x + 6 )( x - 8 )
d) 4x2 + 4x - 15 = ( 4x2 + 4x + 1 ) - 16 = ( 2x + 1 )2 - 42 = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )
e) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
f) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
a) Ta có : x2 - 4x + 3
= x2 - x - 3x + 3
= x(x - 1) - (3x - 3)
= x(x - 1) - 3(x - 1)
= (x - 1) (x - 3)
a) \(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
b) \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+4\right)\)
c) \(x^2-x-6\)
\(=x^2-3x+2x-6\)
\(=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x+2\right)\left(x-3\right)\)
d) \(x^4+1997x^2+1996x+1997\)
\(=x^4+x^2+1996x^2+1996x+1996+1\)
\(=\left(x^4+x^2+1\right)+\left(1996x^2+1996x+1996\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
e) \(x^2-2001\cdot2002\)( hình như sai sai)
a) co sai de ko
b)x3-2x2+4x2-8x+3x-6=x2(x-2)+4x(x-2)+3(x-2)=(x-2)(x2+4x+3)=(x-2)(x+3)(x+1)
c)x3-2x2+2x2-4x-3x+6=x2(x-2)+2x(x-2)-3(x-2)=(x-2)(x2+2x-3)=(x-2)(x+3)(x-1)
d)x3-3x2+x2-3x-2x+6=x2(x-3)+x(x-3)-2(x-3)=(x-3)(x2+x-2)=(x-3)(x+2)(x-1)
c) 2x2 + 10x + 8
= 2x2 + 2x + 8x + 8
= 2x( x + 1) + 8(x + 1)
= 2(x + 1)(x + 4)
d) x2 - 7xy + 10y2
= x2 - 2xy - 5xy + 10y2
= x(x - 2y) - 5y(x - 2y)
= (x - 2y)(x - 5y)
e) x4 + 4x2 - 5
= x4 - x2 + 5x2 - 5
= x2(x2 - 1) + 5(x2 - 1)
= (x - 1)(x + 1)(x2 + 5)
f) x3 - 7x - 6
= x3 - x - 6x - 6
= x(x2 - 1) - 6(x + 1)
= x(x - 1)(x + 1) - 6(x + 1)
= (x + 1)(x - 1)(x - 6)
pn coi kt lại nhé
a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )
c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )
d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )
e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
f) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )
g) x2 + 4x + 3 = x2 + x + 3x + 3 = x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( x + 3 )
h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )